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# Find a function whose ordinate set generates a solid of revolution with volume x2 f(x)

Let be a nonnegative, differentiable function whose graph passes through both points and . For every real number , the ordinate set of on the interval generates a solid of revolution when rotated about the -axis whose volume is given by

Find the formula for the function .

Incomplete.

# Find a function which divides a rectangle into pieces with given properties

Consider a curve whose Cartesian equation is given by , and which passes through the origin. A rectangular region is drawn with one corner at the origin, and the other corner on the curve of the graph of . The curve then divides the rectangle into two pieces and . These two pieces of the rectangle then generate solids of revolution when rotated about the -axis. If the volume of one solid of revolution is always times the volume of the other solid of revolution, find the equation for .

Incomplete.

# Compute the area and volume of solids of revolution of e-2x

Define the function for all . Let

Compute

1. A(t);
2. V(t);
3. W(t);
4. .

1. The area of the ordinate set on is given by the integral,

2. The volume of the solid of revolution obtained by rotating about the -axis is

3. To compute the volume of the solid of revolution obtained by rotating about the -axis we first find as a function of .

Since , the integral is then from to 1 and we have

4. Finally, using parts (c) and (d) we can compute the limit,

# Find a function given the volume of the solid of revolution it generates

Let be a function continuous on an interval . The volume of the solid of revolution obtained by rotating about the -axis on the interval is given by

for every . Find a formula for the function .

Using the formula for the volume of the solid of revolution generated by a function on an interval we know

Now we differentiate both sides of this equation using the fundamental theorem of calculus on the right-hand side,

# Find the slope and area under the graph for a given function

Let

1. Determine the slope of the graph of at the point with -coordinate 1.
2. Find the volume of the solid of revolution formed by rotating the region between the graph of and the interval about the -axis.

1. To take this derivative, using logarithmic differentiation will be easier,

Then differentiating both sides we have,

So, to find the slope at the point with we evaluate,

2. First, the integral to compute the volume of the solid of revolution is,

To evaluate this we use the partial fraction decomposition,

This gives us the equation

Evaluating at , , and we obtain

Therefore, we have

# Compute the volume of the solid of revolution generated by a region bounded by inequalities

Define a region to be the points such that

Sketch this region, and compute the volume of the solid of revolution under the following revolutions:

3. Revolution about the vertical line through the point .
4. Revolution about the horizontal line through the point .

First, the sketch of the region is:

1. We compute the volume from revolving about the -axis as follows,

2. First, we find an equation for in terms of .

Then, we compute the volume as follows from revolving about the -axis as follows,

3. We compute the volume as follows from revolving about the vertical line as follows,

4. We compute the volume as follows from revolving about the horizontal line as follows,

# Compute the volume of a sphere with a cylindrical hole of length 2h removed

Given a solid sphere, drill a cylindrical hole of length through the center of the sphere. Prove that the volume of the resulting ring is , where .

Proof. Let denote the radius of the sphere, and denote the radius of the cylindrical hole. Then the volume of the ring is the volume of the solid of revolution formed by rotating the area between the functions

about the -axis. Since the length of the hole is , we know ; thus, . So, we have,

# Volume of cylindrical hole removed from a sphere

Given a solid sphere of radius , what is the volume of material from a hole of radius through the center of the sphere.

First, the volume of a sphere of radius is given by

Then, the volume of a sphere with a hole drilled in it is the volume of the solid of revolution generated by the region between and from to . Denoting this volume by we then have,

Thus, the volume of the material removed from the sphere by drilling a hole in it is given by

# Find an interval on which the solid of revolution has a particular volume

Let

Find a real number such that the solid of revolution generated by the region between and on the interval has volume .

The volume of the solid of revolution generated by the region between the graphs of and on the interval is,

So, we set this equal to and solve for ,

# Compute the volume of the solid of revolution generated by the region between two functions

Let

Sketch the graph of region between these functions and compute the volume of the solid of revolution generated by revolving this region about the -axis.

The sketch of the region between and is as follows:

We then compute the volume of the solid of revolution as follows.