# volumes of solids of revolution help

Find the volume of the solid formed when the graph of the region bounded by y=e^x y=1 and x=4 is resolved around the x axis.

First, draw the area. There's the horizontal $y=1$ and the vertical $x=4$. The curve $y=e^x$ has value $1$ at $x=0$, so that's where it intersects the bottom line. At $x=4$, it intersects the vertical line at $e^4$.

To get the volume you'll add up the volumes of all of the really thin washers of thickness $dx$, outer diameter of $e^x$, and inner diameter of $1$.

So the volume of the washer element will be the inner hole volume subtracted from the outer disk volume, which is $(\pi (e^x)^2 - \pi 1^2)dx = \pi (e^{2x} - 1)dx.$.

Then the volume of the entire solid just adds all of these washers up via integration:

$$V = \pi \int_0^4 (e^{2x}-1)dx.$$

• Thank you for your help!! I didn't realize I was supposed to use washer and that the intersection point was at e^4. Thanks for your time! – sloth1111 Jan 2 '14 at 22:58
• You're welcome. It took me a bit to wrap my head around this when I was learning it. Do you see how the washers are created when you rotate the strip between $y=1$ and $y=e^x$ about the $x$ axis? I found that once I could visualize how the differential elements were constructed, and what they looked like, things made a lot more sense. – John Jan 2 '14 at 23:14
• now that you mention this I do.. WOW. I'll be sure to visualize now when I solve! – sloth1111 Jan 2 '14 at 23:45

The description of the region is somewhat incomplete. Probably it is intended that the $y$-axis be another bounding line, else the volume is infinite.

Draw the region. Note that when the region is rotated about the $x$-axis, there will be a hole in it, which is a cylinder of radius $1$ and height $4$.

If we rotate the region below $y=e^x$, above the $x$-axis, from $x=0$ to $x=4$, about the $x$-axis, we get volume $$\int_0^4 \pi (e^{x})^2\,dx.$$ This integral is not hard to evaluate, since $(e^x)^2=e^{2x}$. From the definite integral, we must subtract the volume $\pi(1)^2(4)$ of the cylindrical hole.

Another way: Take a cross-section of our solid perpendicular to the $x$-axis, "at" $x$.

We get a circle with a circular hole in it. The outer radius is $e^x$, and the inner radius is $1$. Thus the area of cross-section is $\pi (e^x)^2-\pi(1^2)$. the volume is therefore $$\int_0^4 \left(\pi (e^x)^2-\pi(1^2)\right)\,dx.$$

Another way: We can also use the method of cylindrical shells, though in this case it leads to a more complicated integral. In case you are interested in chasing that one down, we will end up with the integral $$\int_1^{e^4} 2\pi y e^y\,dy.$$

• Thank you so much! The "infinite" volume you stated above was why i got confused. I tried to draw it and didn't know what to integrate. Thank you!!!! – sloth1111 Jan 2 '14 at 22:55
• You are welcome. If we do not bound on the left by the $y$-axis, then in addition there is the region in the second quadrant below $y=1$ and above $y=e^x$. This generates an infinite volume. – André Nicolas Jan 2 '14 at 22:59