Preliminary Mathematics for online MSc programmes in Data AnalyticsUnit 5: Integration in higher dimensions

Integration in higher dimensions

Introduction to integration in higher dimensions and volume of the region

In this Unit we will extend the idea of a definite integral, seen in Unit 3, to double integrals of functions of two variables. The definite integral of a function of one variable represents the area under the curve. Similarly, the double integral of a function of two variables represents the volume of the region between the surface defined by the function and the plane which contains its domain. This can also be used to calculate probabilities when two random variables are involved.

Figure 1

The figure above shows the graph of a function f left parenthesis x comma y right parenthesis over its domain upper R where upper R can be any rectangular region upper R equals left bracket a comma b right bracket times left bracket c comma d right bracket. This notation means that the first variable, x, of the function takes values between a and b while the second variable, y, takes values between c and d. In this case the double integral can be denoted as integral integral Underscript upper R Endscripts f left parenthesis x comma y right parenthesis normal d x normal d y equals integral Subscript c Superscript d Baseline integral Subscript a Superscript b Baseline f left parenthesis x comma y right parenthesis normal d x normal d y period

If f left parenthesis x comma y right parenthesis is continuous on upper R equals left bracket a comma b right bracket times left bracket c comma d right bracket then we can reverse the order of integration. This means that integral Subscript c Superscript d Baseline integral Subscript a Superscript b Baseline f left parenthesis x comma y right parenthesis normal d x normal d y equals integral Subscript a Superscript b Baseline integral Subscript c Superscript d Baseline f left parenthesis x comma y right parenthesis normal d y normal d x period Notice that the inner differential matches up with the limits on the inner integral and similarly for the outer differential and limits. In other words, if the inner differential is normal d y then the limits on the inner integral must be y limits of integration and if the outer differential is normal d y then the limits on the outer integral must be y limits of integration.

In the above example, the integration over x goes from a to b and the integration over y goes from c to d.

Partial integration

Let's look at the inner integral, integral Subscript c Superscript d Baseline f left parenthesis x comma y right parenthesis normal d y, of the last equation. We use the notation integral Subscript c Superscript d Baseline f left parenthesis x comma y right parenthesis normal d y to mean that x is held fixed and we integrate the function f left parenthesis x comma y right parenthesis with respect to y from y equals c to y equals d. This procedure is called partial integration with respect to y.

Now, integral Subscript c Superscript d Baseline f left parenthesis x comma y right parenthesis normal d y is a number that depends on the value of x, so it defines a function of x such as upper A left parenthesis x right parenthesis colon integral Subscript c Superscript d Baseline f left parenthesis x comma y right parenthesis normal d y. If we now integrate the function upper A left parenthesis x right parenthesis with respect to x from x equals a to x equals b, we get integral Subscript a Superscript b Baseline upper A left parenthesis x right parenthesis normal d x equals integral Subscript a Superscript b Baseline left bracket integral Subscript c Superscript d Baseline f left parenthesis x comma y right parenthesis normal d y right bracket normal d x period

The integral on the right side of the equation is called an iterated integral and the brackets are usually omitted. Thus integral Subscript a Superscript b Baseline integral Subscript c Superscript d Baseline f left parenthesis x comma y right parenthesis normal d y normal d x equals integral Subscript a Superscript b Baseline left bracket integral Subscript c Superscript d Baseline f left parenthesis x comma y right parenthesis normal d y right bracket normal d x means that we first integrate with respect to y from c to d and then with respect to x from a to b.

In a similar fashion, the iterated integral integral Subscript c Superscript d Baseline integral Subscript a Superscript b Baseline f left parenthesis x comma y right parenthesis normal d x normal d y equals integral Subscript c Superscript d Baseline left bracket integral Subscript a Superscript b Baseline f left parenthesis x comma y right parenthesis normal d x right bracket normal d y means that we first integrate with respect to x from a to b and then with respect to y from c to d. \newline Notice that in both equations we work from inside out.

Example 1

Let's assume that we have the function f left parenthesis x comma y right parenthesis equals x squared y and we want to evaluate its volume over the region upper R equals left bracket 0 comma 3 right bracket times left bracket 1 comma 2 right bracket.

This means that we want to evaluate the integral integral Subscript 0 Superscript 3 Baseline integral Subscript 1 Superscript 2 Baseline x squared y normal d y normal d x period

As we previously mentioned, we can start by focusing on the inside integral integral Subscript 1 Superscript 2 Baseline x squared y normal d y and regard x as a constant. In that case, we will have StartLayout 1st Row 1st Column integral Subscript 1 Superscript 2 Baseline x squared y normal d y 2nd Column equals x squared integral Subscript 1 Superscript 2 Baseline y normal d y 2nd Row 1st Column Blank 2nd Column equals x squared left bracket StartFraction y squared Over 2 EndFraction right bracket Subscript y equals 1 Superscript y equals 2 Baseline 3rd Row 1st Column Blank 2nd Column equals x squared left parenthesis StartFraction 2 squared Over 2 EndFraction minus StartFraction 1 squared Over 2 EndFraction right parenthesis 4th Row 1st Column Blank 2nd Column equals StartFraction 3 x squared Over 2 EndFraction EndLayout

Thus, the function upper A left parenthesis x right parenthesis in the preceding discussion is equal to StartFraction 3 x squared Over 2 EndFraction.

The only thing left to do right now is to integrate this function with respect to x from x equals 0 to x equals 3. StartLayout 1st Row 1st Column integral Subscript 0 Superscript 3 Baseline integral Subscript 1 Superscript 2 Baseline x squared y normal d y normal d x 2nd Column equals integral Subscript 0 Superscript 3 Baseline left bracket integral Subscript 1 Superscript 2 Baseline x squared y normal d y right bracket normal d x 2nd Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 3 Baseline StartFraction 3 x squared Over 2 EndFraction normal d x 3rd Row 1st Column Blank 2nd Column equals three halves integral Subscript 0 Superscript 3 Baseline x squared normal d x 4th Row 1st Column Blank 2nd Column equals three halves left bracket StartFraction x cubed Over 3 EndFraction right bracket Subscript x equals 0 Superscript x equals 3 Baseline 5th Row 1st Column Blank 2nd Column equals three halves left parenthesis StartFraction 3 cubed Over 3 EndFraction minus StartFraction 0 cubed Over 3 EndFraction right parenthesis 6th Row 1st Column Blank 2nd Column equals StartFraction 27 Over 2 EndFraction period EndLayout That's it. Can you reverse the order of integration and solve the integral, i.e. integral Subscript 1 Superscript 2 Baseline integral Subscript 0 Superscript 3 Baseline x squared y normal d x normal d y

Tasks

Task 1

Evaluate the double integral integral integral Underscript upper R Endscripts left parenthesis x minus 3 y squared right parenthesis normal d x normal d y where upper R equals left brace left parenthesis x comma y right parenthesis vertical bar 0 less than or equals x less than or equals 2 comma 1 less than or equals y less than or equals 2 right brace

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VideoVideo model answersDuration2:05

StartLayout 1st Row 1st Column integral integral Underscript upper R Endscripts x minus 3 y squared normal d x normal d y 2nd Column equals integral Subscript 1 Superscript 2 Baseline integral Subscript 0 Superscript 2 Baseline x minus 3 y squared normal d x normal d y 2nd Row 1st Column Blank 2nd Column equals integral Subscript 1 Superscript 2 Baseline left bracket StartFraction x squared Over 2 EndFraction minus 3 x y squared right bracket Subscript x equals 0 Superscript 2 Baseline normal d y 3rd Row 1st Column Blank 2nd Column equals integral Subscript 1 Superscript 2 Baseline 2 minus 6 y squared normal d y 4th Row 1st Column Blank 2nd Column equals left bracket 2 y minus 2 y cubed right bracket Subscript y equals 1 Superscript 2 Baseline 5th Row 1st Column Blank 2nd Column equals negative 12 EndLayout

Task 2

Find integral Subscript 0 Superscript 5 Baseline f left parenthesis x comma y right parenthesis normal d x and integral Subscript 0 Superscript 1 Baseline f left parenthesis x comma y right parenthesis normal d y for

(a) f left parenthesis x comma y right parenthesis equals 12 x squared y cubed

(b) f left parenthesis x comma y right parenthesis equals y plus x e Superscript y

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(a) 500 y cubed and 3 x squared

(b) 5 y plus StartFraction 25 Over 2 EndFraction e Superscript y and one half plus e x minus x

Task 3

Calculate the iterated integrals:

(a) integral Subscript 1 Superscript 4 Baseline integral Subscript 0 Superscript 2 Baseline left parenthesis 6 x squared y minus 2 x right parenthesis normal d y normal d x

(b) integral Subscript 1 Superscript 4 Baseline integral Subscript 1 Superscript 2 Baseline left parenthesis StartFraction x Over y EndFraction plus StartFraction y Over x EndFraction right parenthesis normal d y normal d x

(c) integral Subscript 0 Superscript 1 Baseline integral Subscript 0 Superscript 1 Baseline w left parenthesis z plus w squared right parenthesis Superscript 4 normal d z normal d w

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Task 4

Calculate the following double integrals:

(a) integral integral Underscript upper R Endscripts StartFraction x y squared Over x squared plus 1 EndFraction normal d x normal d y where upper R equals left brace left parenthesis x comma y right parenthesis vertical bar 0 less than or equals x less than or equals 1 comma negative 3 less than or equals y less than or equals 3 right brace

(b) integral integral Underscript upper R Endscripts y e Superscript minus x y normal d x normal d y where upper R equals left brace left parenthesis x comma y right parenthesis vertical bar 0 less than or equals x less than or equals 2 comma 0 less than or equals y less than or equals 3 right brace

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VideoVideo model answers for part (a)Duration2:47

(a) 9 log 2

(b) one half left parenthesis e Superscript negative 6 Baseline plus 5 right parenthesis

Self help

Integration over general regions

For single integrals, the regions over which we integrate is always an interval. In the previous section we looked at double integrals which have to be integrated over rectangular regions. The problem with this is that most of the regions are not rectangular so we need to now look at the following double integral

Figure 2

integral integral Underscript upper R Endscripts f left parenthesis x comma y right parenthesis normal d x normal d y equals integral integral Underscript upper R Endscripts f left parenthesis x comma y right parenthesis normal d y normal d x where upper R is a non-rectangular region. The figure above shows the graph of a function f left parenthesis x comma y right parenthesis over its domain upper R where upper R is now the non-rectangular region in green. These integrals are unfortunately more complex to solve compared to the integrals over rectangular regions. The reason being it could be the case that the left hand side integral, on the previous equation, is difficult to solve while the one on the right-hand side is easy. This means that we would have to change the order of integration and put the correct limits of integration for the new integral.

Changing the order of integration is slightly tricky because its hard to write down a specific algorithm for the procedure. The easiest way to accomplish the task is through drawing a picture of the region upper R. From the picture, we can determine the corners and edges of the region upper R, which is what we need to work out (i.e. the limits of integration). Let's look at an example.

Example 2

Assume that we want to calculate the integral integral Subscript 0 Superscript 1 Baseline integral Subscript x Superscript 1 Baseline e Superscript y squared Baseline normal d y normal d x period Evaluating the inner integral with respect to y will be very difficult as there is no anti-derivative of e Superscript y squared (i.e. a function that when we differentiate it we will get e Superscript y squared). But, if we manage to change the order of integration, we can integrate with respect to x first which is doable. And, it turns out that the integral with respect to y also becomes possible after we finish integrating with respect to x.

According to the limits of the integral, the region upper R can be described as

StartLayout 1st Row 1st Column 0 2nd Column less than or equals x less than or equals 1 2nd Row 1st Column x 2nd Column less than or equals y less than or equals 1 EndLayout

The figure below shows the aforementioned region.

Figure 3

Since we can also describe the region upper R by

StartLayout 1st Row 1st Column 0 2nd Column less than or equals y less than or equals 1 2nd Row 1st Column 0 2nd Column less than or equals x less than or equals y EndLayout

the integral with the order changed is integral Subscript 0 Superscript 1 Baseline integral Subscript 0 Superscript y Baseline e Superscript y squared Baseline normal d x normal d y period

Is it easier to solve this integral instead of the one we started with? Let's check this by starting with the inner integral first and integrate with respect to x (ending up with a function upper A left parenthesis y right parenthesis).

StartLayout 1st Row 1st Column integral Subscript 0 Superscript y Baseline e Superscript y squared Baseline normal d x 2nd Column equals e Superscript y squared Baseline integral Subscript 0 Superscript y Baseline 1 normal d x 2nd Row 1st Column Blank 2nd Column equals e Superscript y squared Baseline left bracket x right bracket Subscript x equals 0 Superscript x equals y Baseline 3rd Row 1st Column Blank 2nd Column equals e Superscript y squared Baseline left parenthesis y minus 0 right parenthesis 4th Row 1st Column Blank 2nd Column equals e Superscript y squared Baseline y EndLayout

Thus, the function upper A left parenthesis y right parenthesis equals e Superscript y squared Baseline y.

The only thing left to do right now is to integrate the function with respect to y from y equals 0 to y equals 1.

StartLayout 1st Row 1st Column integral Subscript 0 Superscript 1 Baseline integral Subscript 0 Superscript y Baseline e Superscript y squared Baseline normal d x normal d y 2nd Column equals integral Subscript 0 Superscript 1 Baseline left bracket integral Subscript 0 Superscript y Baseline e Superscript y squared Baseline normal d x right bracket normal d y 2nd Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 1 Baseline e Superscript y squared Baseline y normal d y EndLayout

This can be integrated by using the substitution u equals y squared. Using this substitution we will also have normal d u equals 2 y normal d y. We also need to work out the limits of the new integral. When y equals 0 we have u equals 0 squared equals 0 and when y equals 1 we get u equals 1 squared equals 1. We now have

StartLayout 1st Row 1st Column integral Subscript 0 Superscript 1 Baseline e Superscript y squared Baseline y normal d y 2nd Column equals integral Subscript 0 Superscript 1 Baseline e Superscript u Baseline StartFraction CrossOut y EndCrossOut Over 2 CrossOut y EndCrossOut EndFraction normal d u 2nd Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 1 Baseline e Superscript u Baseline one half normal d u 3rd Row 1st Column Blank 2nd Column equals one half integral Subscript 0 Superscript 1 Baseline e Superscript u Baseline normal d u 4th Row 1st Column Blank 2nd Column equals one half left bracket e Superscript u Baseline right bracket Subscript u equals 0 Superscript u equals 1 Baseline 5th Row 1st Column Blank 2nd Column equals one half left parenthesis e Superscript 1 Baseline minus e Superscript 0 Baseline right parenthesis 6th Row 1st Column Blank 2nd Column equals one half left parenthesis e minus 1 right parenthesis EndLayout

Example 3

The focus of this example is on the limits of integration so there is no need to specify the function f left parenthesis x comma y right parenthesis. The procedure does not depend on the identity of f.

Assume that we have a function f left parenthesis x comma y right parenthesis and we want to calculate the integral integral Subscript 0 Superscript 1 Baseline integral Subscript 1 Superscript e Superscript y Baseline Baseline f left parenthesis x comma y right parenthesis normal d x normal d y period

We can see that on the previous integral, the integration order is normal d x normal d y. As we previously mentioned, this corresponds to first integrating with respect to x from x equals 0 to x equals e Superscript y, and afterwards integrating with respect to y from y equals 0 to y equals 1. Our task is to change the order of integration to be normal d y normal d x.

We begin by transforming the limits of integration into the domain upper R. The limits of the outer normal d y integral mean that 0 less than or equals y less than or equals 1, and the limits on the inner normal d x integral mean that for each value of y the range of x is 1 less than or equals x less than or equals e Superscript y. The region upper R is shown in the figure below.

Figure 4

The range of y over the region is from 0 to 1, as indicated by the gray bar to the left of the figure. The horizontal hashing within the figure indicates the range of x for each value of y, beginning at the left edge x equals 1 (blue line) and ending at the right curve edge x equals e Superscript y (red curve).

We have also labelled all the corners of the region. The upper-right corner is the intersection of the line y equals 1 with the curve x equals e Superscript y. Therefore, the value of x at this corner must be e Superscript 1 Baseline equals e, and the point is left parenthesis e comma 1 right parenthesis.

To change the order of integration, we need to write an integral with order normal d y normal d x. This means that x should be the variable of the outer integral. Its limits must be constant and correspond to the total range of x over the region upper R. The range of x is 1 less than or equals x less than or equals e Superscript 1, as indicated by the gray bar below the region in the figure below.

Figure 5

Since y will be the variable for the inner integration, we need to integrate with respect to y first. The vertical hashing indicates how, for each value of x, we will integrate from the lower boundary (red curve) to the upper boundary (purple line). These two boundaries determine the range of y. Since we can rewrite the equation x equals e Superscript y for the red curve as y equals log left brace x right brace, the range of y is log left brace x right brace less than or equals y less than or equals 1. (Note that this indicates the natural logarithm since the base of this logarithmic function is e. This means that we can write it as ln left brace x right brace instead)

In summary, the region upper R can be described not only by StartLayout 1st Row 1st Column 0 2nd Column less than or equals y less than or equals 1 2nd Row 1st Column 1 2nd Column less than or equals x less than or equals e Superscript y EndLayout

as it was for the original normal d x normal d y integral, but also by

StartLayout 1st Row 1st Column 1 2nd Column less than or equals x less than or equals e 2nd Row 1st Column log left brace x right brace 2nd Column less than or equals y less than or equals 1 EndLayout

which is the description we need for the new normal d y normal d x integral. We can now write that

integral Subscript 0 Superscript 1 Baseline integral Subscript 1 Superscript e Superscript y Baseline Baseline f left parenthesis x comma y right parenthesis normal d x normal d y equals integral Subscript 1 Superscript e Baseline integral Subscript log left brace x right brace Superscript 1 Baseline f left parenthesis x comma y right parenthesis normal d y normal d x

In general, to solve these double integrals we need to remember some points:

  1. After we have set up the integral, visualise the region that is encoded by the bounds in the integral. If in case, that region does not look as specified in the problem, double check the entire setup.

  2. Make sure that the bounds of the outer integral are numbers or constants.

  3. Make sure that the bounds of the inner integral depend only on the integration variable of the outer integral. In particular, the bounds must not depend on the integration variable of the inner integral.

Tasks

Task 5

Calculate the volume under the surface z equals 3 plus x squared minus 2 y over the region upper R defined by \newline 0 less than or equals x less than or equals 1 and negative x less than or equals y less than or equals x.

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The volume V is the double integral of 3 plus x squared minus 2 y StartLayout 1st Row 1st Column upper V 2nd Column equals integral integral Underscript upper R Endscripts left parenthesis 3 plus x squared minus 2 y right parenthesis normal d x normal d y 2nd Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 1 Baseline integral Subscript negative x Superscript x Baseline left parenthesis 3 plus x squared minus 2 y right parenthesis normal d y normal d x 3rd Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 1 Baseline left bracket 3 y plus x squared y minus y squared right bracket Subscript y equals negative x Superscript y equals x Baseline normal d x 4th Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 1 Baseline left parenthesis 6 x plus 2 x cubed right parenthesis normal d x 5th Row 1st Column Blank 2nd Column equals left bracket StartFraction 6 x squared Over 2 EndFraction plus StartFraction 2 x Superscript 4 Baseline Over 4 EndFraction right bracket Subscript x equals 0 Superscript x equals 1 Baseline 6th Row 1st Column Blank 2nd Column equals seven halves period EndLayout

Task 6

Evaluate the following integrals by first reversing the order of integration.

(a) integral Subscript 0 Superscript 3 Baseline integral Subscript x squared Superscript 9 Baseline x cubed e Superscript y cubed normal d y normal d x (b) integral Subscript 0 Superscript 8 Baseline integral Subscript RootIndex 3 StartRoot y EndRoot Superscript 2 Baseline StartRoot x Superscript 4 Baseline plus 1 EndRoot normal d x normal d y

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(a)

We need to compute the integral integral Subscript 0 Superscript 3 Baseline integral Subscript x squared Superscript 9 Baseline x cubed e Superscript y cubed normal d y normal d x.

Notice that if we try to integrate with respect to y we can't calculate the integral because we can't integrate e Superscript y cubed. Let's hope that if we reverse the order, the integrals will be easier to solve.

Again, when we say that we're going to reverse the order of integration this means that we want to integrate with respect to x first and then y. Note as well that we can't just interchange the integrals, keeping the original limits, and be done with it. This would not fix our original problem and in order to integrate with respect to x we can’t have x's in the limits of the integrals. Even if we ignored that the answer would not be a constant as it should be.

So, let’s see how we reverse the order of integration. The best way to reverse the order of integration is to first sketch the region given by the original limits of integration.

Figure 6

From the integral we see that the inequalities that define this region are,

StartLayout 1st Row 1st Column 0 2nd Column less than or equals x less than or equals 3 2nd Row 1st Column x squared 2nd Column less than or equals y less than or equals 9 EndLayout

These inequalities tell us that we want the region with y equals x squared on the lower boundary and y equals 9 on the upper boundary that lies between x equals 0 and x equals 3.

Since we want to integrate with respect to x first we will need to determine limits of x (probably in terms of y) and then get the limits on the y's. These are:

StartLayout 1st Row 1st Column 0 2nd Column less than or equals x less than or equals StartRoot y EndRoot 2nd Row 1st Column 0 2nd Column less than or equals y less than or equals 9 EndLayout

Any horizontal line drawn in this region will start at x equals 0 and end at x equals StartRoot y EndRoot and so these are the limits on the x's and the range of y's for the region is 0 to 9.

Reversing the order of integration we now have that integral Subscript 0 Superscript 3 Baseline integral Subscript x squared Superscript 9 Baseline x cubed e Superscript y cubed Baseline normal d y normal d x equals integral Subscript 0 Superscript 9 Baseline integral Subscript 0 Superscript StartRoot y EndRoot Baseline x cubed e Superscript y cubed Baseline normal d x normal d y period so we can now start calculating the integral on the right-hand side. StartLayout 1st Row 1st Column integral Subscript 0 Superscript 9 Baseline left bracket integral Subscript 0 Superscript StartRoot y EndRoot Baseline x cubed e Superscript y cubed Baseline normal d x right bracket normal d y 2nd Column equals integral Subscript 0 Superscript 9 Baseline e Superscript y cubed Baseline left bracket StartFraction x Superscript 4 Baseline Over 4 EndFraction right bracket Subscript x equals 0 Superscript x equals StartRoot y EndRoot Baseline normal d y 2nd Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 9 Baseline e Superscript y cubed Baseline StartFraction y squared Over 4 EndFraction normal d y EndLayout Although this looks tricky, having y squared multiplied with e Superscript y cubed makes life easier. Why?

If we substitute u equals y cubed we get normal d u equals 3 y squared normal d y. The new lower limit will now be u equals 0 cubed equals 0 and the new upper limit will be u equals 9 cubed equals 729. We now have StartLayout 1st Row 1st Column integral Subscript 0 Superscript 9 Baseline one fourth e Superscript y cubed Baseline y squared normal d y 2nd Column equals integral Subscript 0 Superscript 729 Baseline one fourth e Superscript u Baseline StartFraction CrossOut y squared EndCrossOut Over 3 CrossOut y squared EndCrossOut EndFraction normal d u 2nd Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 729 Baseline one twelfth e Superscript u Baseline normal d u 3rd Row 1st Column Blank 2nd Column equals one twelfth left parenthesis e Superscript 729 Baseline minus 1 right parenthesis EndLayout

(b)

In this part we compute integral Subscript 0 Superscript 8 Baseline integral Subscript RootIndex 3 StartRoot y EndRoot Superscript 2 Baseline StartRoot x Superscript 4 Baseline plus 1 EndRoot normal d x normal d y.

As with the first integral we cannot calculate it by integrating with respect to x first so we'll hope that by reversing the order of integration we will get something that we can integrate. Here are the limits for the variables that we get from this integral:

StartLayout 1st Row 1st Column RootIndex 3 StartRoot y EndRoot 2nd Column less than or equals x less than or equals 2 2nd Row 1st Column 0 2nd Column less than or equals y less than or equals 8 EndLayout

and here is a sketch for the region

Figure 7

If we now reverse the order of integration we will get the following region:

StartLayout 1st Row 1st Column 0 2nd Column less than or equals x less than or equals 2 2nd Row 1st Column 0 2nd Column less than or equals y less than or equals x cubed EndLayout

The integral is then, integral Subscript 0 Superscript 8 Baseline integral Subscript RootIndex 3 StartRoot y EndRoot Superscript 2 Baseline StartRoot x Superscript 4 Baseline plus 1 EndRoot normal d x normal d y equals integral Subscript 0 Superscript 2 Baseline integral Subscript 0 Superscript x cubed Baseline StartRoot x Superscript 4 Baseline plus 1 EndRoot normal d y normal d x and StartLayout 1st Row 1st Column integral Subscript 0 Superscript 2 Baseline integral Subscript 0 Superscript x cubed Baseline StartRoot x Superscript 4 Baseline plus 1 EndRoot normal d y normal d x 2nd Column equals integral Subscript 0 Superscript 2 Baseline left bracket integral Subscript 0 Superscript x cubed Baseline StartRoot x Superscript 4 Baseline plus 1 EndRoot normal d y right bracket normal d x 2nd Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 2 Baseline left bracket y StartRoot x Superscript 4 Baseline plus 1 EndRoot right bracket Subscript y equals 0 Superscript y equals x cubed Baseline normal d x 3rd Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 2 Baseline x cubed StartRoot x Superscript 4 Baseline plus 1 EndRoot normal d x EndLayout

A tricky integral until we notice that we can use the substitution method. If we substitute u equals x Superscript 4 Baseline plus 1 we will get normal d u equals 4 x cubed normal d y. The new lower limit will now be u equals 0 Superscript 4 Baseline plus 1 equals 1 and the new upper limit will be u equals 2 Superscript 4 Baseline plus 1 equals 17. We now have

StartLayout 1st Row 1st Column integral Subscript 0 Superscript 2 Baseline x cubed StartRoot x Superscript 4 Baseline plus 1 EndRoot normal d x 2nd Column equals integral Subscript 0 Superscript 64 Baseline CrossOut x cubed EndCrossOut StartRoot u EndRoot StartFraction 1 Over CrossOut 4 x cubed EndCrossOut EndFraction normal d u 2nd Row 1st Column Blank 2nd Column equals integral Subscript 0 Superscript 64 Baseline one fourth StartRoot u EndRoot normal d u 3rd Row 1st Column Blank 2nd Column equals one fourth left bracket StartFraction u Superscript 3 divided by 2 Baseline Over three halves EndFraction right bracket Subscript u equals 0 Superscript u equals 64 Baseline 4th Row 1st Column Blank 2nd Column equals one sixth left parenthesis 17 Superscript 3 divided by 2 Baseline minus 1 right parenthesis EndLayout

Self help

VideoDouble integrals over a non-rectangular region (YouTube video)Duration4:20 VideoChanging order of integration in a double integral (YouTube video)Duration3:40