# Calculus 3 : Double Integration over General Regions

## Example Questions

### Example Question #2191 : Calculus 3

Evaluate the following integral:

Explanation:

First, you must evaluate the integral with respect to z. Using the rules for integration, we get  evaluated from  to . The result is . This becomes , evaluated from  to . The final answer is .

### Example Question #2192 : Calculus 3

Evaluate:

Explanation:

To evaluate the iterated integral, we start with the innermost integral, evaluated with respect to x:

The integral was found using the following rule:

Now, we evaluate the last remaining integral, using our answer above from the previous integral as our integrand:

The integral was found using the following rule:

### Example Question #2193 : Calculus 3

Evaluate the double integral

Explanation:

To evaluate the double integral, compute the inside integral first.

### Example Question #2201 : Calculus 3

Evaluate the double integral

Explanation:

aTo evaluate the double integral, compute the inside integral first.

### Example Question #2202 : Calculus 3

Evaluate the double integral

Explanation:

To evaluate the double integral, compute the inside integral first.

### Example Question #11 : Double Integration Over General Regions

Integrate:

Explanation:

To perform the iterated integration, we must work from inside outwards. To start we perform the following integration:

This becomes the integrand for the outermost integral:

### Example Question #682 : Multiple Integration

Explanation:

To perform the iterated integral, we work from inside outwards.

The first integral we perform is

This becomes the integrand for the outermost integral,

### Example Question #683 : Multiple Integration

Solve:

Explanation:

To evaluate the iterated integral, we must work from inside outward.

The first integral we evaluate is

This becomes the integrand for the outermost integral.

The final integral we evaluate is

### Example Question #11 : Double Integration Over General Regions

Let  be any continuous, one-to-one function over the interval , with on  and . Select all integrals which correctly define the indicated area under the curve.

1 and 3

1 3 and 4

1

1 and 4

3 and 4

1 and 3

Explanation:

No explanation necessary. This is the familiar integral for computing the area under the curve  over

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This integral does not give the area, in fact, it will result in a function of  which certainly cannot be interpreted as the area of corresponding to a specific interval

Now carry out the integration with respect to  we obtain,

Certainly this is not the area of the area under . One obvious problem is that the integral gave back a function of  and no real number.

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This equation will reduce down to the first option, option , after we carry out the integration with respect to .

Therefore, option  is a valid representation for the area.

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The problem with this choice can be seen immediately. Clearly the first integration in will simply be the area under the curve on , but the second will now multiply the area by

could only be true if . But because none of the listed characteristics of our function imply this, we cannot assume it. In fact, the problem stated that  is any function that satisfies the stated conditions. Certainly  is not true for just any function.

### Example Question #685 : Multiple Integration

Evaluate the following iterated integrals:

Explanation:

When evaluating double integrals, work from the inside out: that is, evaluate the integrand with respect to the first variable listed by the differential operators, and then evaluate the result with respect to the second variable listed by the differential operators.

Here, we have the order of integration specified by ; hence, we evaluate the double integrals with respect to  first, and then integrate the result with respect to , as shown: