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.**

This one was a little more tricky than the previous problem, but after the initial setup, the overall approach was almost identical.

If A is the region in beneath the horizontal line y = f(x) and above the function y = f(t) from t = 0 to t = x, then the volume swept out by rotating this region around the t-axis is:

And if B is the region beneath the function y = f(t) and above the t-axis from t = 0 to t = x, then the volume swept out by rotating this region around the t-axis is:

But we know that the volume swept out by A is n times the volume swept out by B, so we have

Or in other words

Evaluating the left side gives us

Differentiating both sides with respect to x gives us

Rearranging terms gives us logarithmic derivatives on both sides

Integrating both sides gives us:

Taking the exponential of both sides gives us:

Where:

Now, the back of the book has it such that the answer could also be

It appears that we could reach this answer by switching A and B in our initial setup, such that B is n times A, and proceeding the same way as above.

I forgot to multiply by pi in the integrals for the volumes. But, since the factors of pi were always on both sides, they end up cancelling out and the final answer remains unaffected… luckily for me.