### All Calculus 3 Resources

## Example Questions

### Example Question #2191 : Calculus 3

Evaluate the following integral:

**Possible Answers:**

**Correct answer:**

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:

**Possible Answers:**

**Correct answer:**

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

**Possible Answers:**

**Correct answer:**

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

### Example Question #2201 : Calculus 3

Evaluate the double integral

**Possible Answers:**

**Correct answer:**

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

### Example Question #2202 : Calculus 3

Evaluate the double integral

**Possible Answers:**

**Correct answer:**

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

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

Integrate:

**Possible Answers:**

**Correct answer:**

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

**Possible Answers:**

**Correct answer:**

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:

**Possible Answers:**

**Correct answer:**

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.

**Possible Answers:**

1 and 3

1 3 and 4

1

1 and 4

3 and 4

**Correct answer:**

1 and 3

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:

**Possible Answers:**

**Correct answer:**

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:

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