All AP Calculus BC Resources
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:
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?
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.
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:
Example Question #71 : 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 .
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 #72 : 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.
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 #73 : Integrals
Determine the volume of a solid created by rotating the curve and the line by revolving around the -axis.
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 #74 : Integrals
What is the volume of the solid formed when the line is rotated around the -axis from to ?
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 #75 : Integrals
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.
The formula for the volume is given as
where and the bounds on the integral come from the interval .
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
Example Question #76 : Integrals
Using the method of cylindrical disks, find the volume of the region revolved around the x-axis of the graph of
on the interval
The formula for volume of the region revolved around the x-axis is given as
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
Example Question #77 : Integrals
Find the volume of the solid generated by rotating about the y-axis the region under the curve , from to .
None of the other answers
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