I have a volumes problem.
Find the volume obtained by rotating this region about the line y=3
The two curves are y=e^x and y=cos(2 Pi x) from x=0 to x=1
I tried setting up the problem but I'm not sure if it's correct. This is what I got: 2pi integral from 0 to 1 (3-x)(cos(2pix)-e^x)dx
Is this correct?

Question

I have a volumes problem. 
Find the volume obtained by rotating this region about the line y=3
The two curves are y=e^x and y=cos(2 Pi x) from x=0 to x=1
 I tried setting up the problem but I'm not sure if it's correct. This is what I got: 2pi integral from 0 to 1 (3-x)(cos(2pix)-e^x)dx
Is this correct?

anonymous · Answer

Slight mistake. Notice that $e^x$ is strictly increasing, while $\cos2\pi x$ is bounded between -1 and 1.

Take the endpoints: For $x=0$, you have $e^0=\cos0=1$, but for $x=1$ you have $e^1=e>1=\cos2\pi$, which means the exponential curve is above the trigonometric curve.

The general setup of your integral is correct, but if you were to evaluate it you would get the negative of the correct answer.

anonymous · Answer

I ended up changing my formula and I got pi integral 0 to 1 (3-cos(2 pi x))^2-(e^x-3)^2 
is this correct?

anonymous · Answer

The washer method for this problem is quite a bit more involved than the shell method, I'm afraid.
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