AP Calculus BC : Volume of Cross Sections and Area of Region

Study concepts, example questions & explanations for AP Calculus BC

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Example Questions

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Example Question #1 : Volume Of Cross Sections And Area Of Region

Determine the volume of the solid obtained by rotating the region with the following bounds about the x-axis:

Possible Answers:

Correct answer:

Explanation:

From calculus, we know the volume of an irregular solid can be determined by evaluating the following integral:

Where A(x) is an equation for the cross-sectional area of the solid at any point x. We know our bounds for the integral are x=1 and x=4, as given in the problem, so now all we need is to find the expression A(x) for the area of our solid.

From the given bounds, we know our unrotated region is bounded by the x-axis (y=0) at the bottom, and by the line y=x^2-4x+5 at the top. Because we are rotating about the x-axis, we know that the radius of our solid at any point x is just the distance y=x^2-4x+5. Now that we have a function that describes the radius of the solid at any point x, we can plug the function into the formula for the area of a circle to give us an expression for the cross-sectional area of our solid at any point:

We now have our equation for the cross-sectional area of the solid, which we can integrate from x=1 to x=4 to find its volume:

Example Question #2 : Volume Of Cross Sections And Area Of Region

Suppose the functions , and  form a closed region. Rotate this region across the x-axis. What is the volume?

Possible Answers:

Correct answer:

Explanation:

Write the formula for cylindrical shells, where  is the shell radius and  is the shell height.

Determine the shell radius.

Determine the shell height.  This is done by subtracting the right curve, , with the left curve, .

Find the intersection of  and  to determine the y-bounds of the integral.

The bounds will be from 0 to 2.  Substitute all the givens into the formula and evaluate the integral.

Example Question #3 : Volume Of Cross Sections And Area Of Region

Find the volume of the solid generated by revolving the region bounded by  and the -axis in the first quadrant about the -axis.

Possible Answers:

Correct answer:

Explanation:

Since we are revolving a region of interest around a horizontal line , we need to express the inner and outer radii in terms of x.

Recall the formula:

The outer radius is  and the inner radius is . The x-limits of the region are between  and . So the volume set-up is:

Using trigonometric identities, we know that:    

Hence:

 

 

 

 

Example Question #8 : Application Of Integrals

A man fills up a cup of water by leaving it outside during a rainstorm. The rate at which the height of the cup changes is equal to . What is the height of water at ? Assume the cup is empty at .

Possible Answers:

Correct answer:

Explanation:

The rate at which the height changes is , which means .

To find the height after nine seconds, we need to integrate to get .

We can multiply both sides by  to get  and then integrate both sides.

This gives us 

.

Since the cup is empty at , so .

This means . No units were given in the problem, so leaving the answer unitless is acceptable.

Example Question #9 : Application Of Integrals

Approximate the volume of a solid in the first quadrant revolved about the y-axis and bounded by the functions:   and .  Round the volume to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

Write the washer's method.

Set the equations equal to each other to determine the bounds.

The bounds are from 0 to 3.

Determine the big and small radius.  Rewrite the equations so that they are in terms of y.

Set up the integral and solve for the volume.

The volume to the nearest integer is:  

Example Question #4 : Volume Of Cross Sections And Area Of Region

Determine the volume of a solid created by rotating the curve  and the line  by revolving around the -axis.

Possible Answers:

Correct answer:

Explanation:

Write the volume formula for cylindrical shells.

The shell radius is .

The shell height is the function in terms of .  Rewrite that equation.

The bounds lie on the y-axis since the thickness variable is .  This is from 0 to 1, since the intersection of the line  and  is at .  

Substitute all the values and solve for the volume.

Example Question #4 : Volume Of A Solid

What is the volume of the solid formed when the line  is rotated around the -axis from  to ?

Possible Answers:

Correct answer:

Explanation:

To rotate a curve around the y-axis, first convert the function so that y is the independent variable by solving  for x. This leads to the function 

We'll also need to convert the endpoints of the interval to y-values. Note that when , and when  Therefore, the the interval being rotated is from .

The disk method is best in this case. The general formula for the disk method is 

, where V is volume,  are the endpoints of the interval, and  the function being rotated. 

 

Substuting the function and endpoints from the problem at hand leads to the integral

.

To evaluate this integral, you must know the power rule. Recall that the power rule is 

.

Example Question #5 : Volume Of A Solid

Find the volume of the solid generated when the function 

 

is revolved around the x-axis on the interval .

Hint: Use the method of cylindrical disks.

Possible Answers:

 units cubed

 units cubed

 units cubed

 units cubed

Correct answer:

 units cubed

Explanation:

The formula for the volume is given as

where  and the bounds on the integral come from the interval .

As such, 

When taking the integral, we will use the inverse power rule which states

Applying this rule we get

And by the corollary of the First Fundamental Theorem of Calculus

As such,

 units cubed

Example Question #5 : Volume Of Cross Sections And Area Of Region

Using the method of cylindrical disks, find the volume of the region revolved around the x-axis of the graph of 

on the interval 

Possible Answers:

 units cubed

 units cubed

 units cubed

 units cubed

Correct answer:

 units cubed

Explanation:

The formula for volume of the region revolved around the x-axis is given as 

where 

As such

When taking the integral, we use the inverse power rule which states

Applying this rule term by term we get

And by the corollary of the Fundamental Theorem of Calculus 

As such the volume is

 units cubed

Example Question #6 : Volume Of Cross Sections And Area Of Region

Find the volume of the solid generated by rotating about the y-axis the region under the curve , from  to .

Possible Answers:

None of the other answers

Correct answer:

Explanation:

Since we are revolving a function of  around the y-axis, we will use the method of cylindrical shells to find the volume.

 

Using the formula for cylindrical shells, we have

 

.

 

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