Finding the volume of a solid rotated around the line x = -1

[tarynt1]

Consider the plane region bounded by y = e^x, x = 4, and y = 1.

Revolve this region about the line y = -1 and find the resulting volume of the solid.

I understand most of the disk and washer method for finding volumes, but revolving the solid around y = -1 confuses me; I don't understand how to set up the integral.

The answer is supposed to be approximately 4979.975, but I would just like help on setting up the integral.

 

[galactus]

You can use washers.

\(\displaystyle \L\\{\pi}\int_{0}^{4}\left[(e^{x}-(-1))^{2}-(1-(-1))^{2}\right]dx\)

\(\displaystyle \L\\{\pi}\int_{0}^{4}\left[e^{2x}+2e^{x}-3\right]dx\)

See what you must do with the -1?.

Now, let's see about the shells method:

\(\displaystyle \L\\2{\pi}\int_{1}^{e^{4}}\left[(y+1)(4-ln(y))\right]dy\)

Evaluate both integrals. Do you get the same thing?.

 

[tarynt1]

Makes sense, thank you!