Question

What integral will find the volume of the solid that is formed when the region bounded by the graphs of y = e^x, x = 2, and y = 1 is revolved around the line y = −1?

Answer 1

So umm, let's see...

Answer 2

If we take a slice with "thickness" dx...
and spin it around...

Answer 3

then we get this washer shape.
Let's see if we can properly label the dimensions of this washer.

Answer 4

Our outer radial length is\[\large\rm R=e^x+1\]
While our inner length measures\[\large\rm r=2\]

Answer 5

To find the volume of our washer,
we'll find the surface area (subtracting the hole in the middle): \(\large\rm \pi(R^2-r^2)\)
(It's basically just big circle area minus little circle area.)

and then multiplying that by the "thickness": \(\large\rm \pi(R^2-r^2)dx\)

Answer 6

\[\large\rm dV=\pi[R^2-r^2]dx\]Filling in our radial values,\[\large\rm dV=\pi\left[(e^x+1)^2-2^2\right]dx\]

Answer 7

And then, integrate over the region "adding up" all these washers.
\[\large\rm V=\int\limits dV\]
\[\large\rm V=\int\limits_{x=0}^1\pi\left[(e^x+1)^2-4\right]dx\]

Answer 8

Woops, from x=0 to x=2.
I had to look back at the picture again :p