All AP Calculus BC Resources
Example Questions
Example Question #2 : Volume Of A Solid
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 #1 : Volume Of Cross Sections And Area Of Region
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 #4 : Volume Of A Solid
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 #1 : Volume Of A Solid
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 #1 : 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.
units cubed
units cubed
units cubed
units cubed
units cubed
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 #1 : 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
units cubed
units cubed
units cubed
units cubed
units cubed
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 #11 : Application Of 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
.
Example Question #181 : Ap Calculus Bc
Find the area bound by the curve of g(t), the x and y axes, and the line
Find the area bound by the curve of g(t), the x and y axes, and the line
We are asked to find the area under a curve. This sounds like a job for an integral.
We need to integrate our function from 0 to . These will be our limits of integration. With that in mind, our integral look like this.
Now, we need to recall the rules for integrating sine and cosine.
With these rules in mind, we can integrate our original function to get:
Now, we just need to evaluate our new function at the given limits. Let's start with 0
And our upper limit...
Now, we need to take the difference between our limits:
So, our answer is 2.59
Example Question #171 : Integrals
Determine the length of the following function between
In order to begin the problem, we must first remember the formula for finding the arc length of a function along any given interval:
where ds is given by the equation below:
We can see from our equation for ds that we must find the derivative of our function, which in our case is dv/dt instead of dy/dx, so we begin by differentiating our function v(t) with respect to t:
Now we can plug this into the given equation to find ds:
Our last step is to plug our value for ds into the equation for arc length, which we can see only involves integrating ds. The interval along which the problem asks for the length of the function gives us our limits of integration, so we simply integrate ds from t=1 to t=4:
Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve
In physics, the work done on an object is equal to the integral of the force on that object dotted with its displacent.
This looks like ( is work, is force, and is the infinitesimally small displacement vector). For a force whose direction is the line of motion, the equation becomes .
If the force on an object as a function of displacement is , what is the work as a function of displacement ? Assume and the force is in the direction of the object's motion.
Not enough information
, so .
Both the terms of the force are power terms in the form , which have the integral , so the integral of the force is .
We know
.
This means
.