Volume of a solid of revolution around the y-axis (def. integration)

[greg_rack]

Homework Statement

Calculate the volume of the solid generated by rotating around the y-axis the plane region delimited by curves:
$y=e^x$
$x=0$
$x=1$
$y=0$

Relevant Equations

Definite integrals definition

First, I calculated the inverse of $y=e^x$ since we're talking about y-axis rotations, which is of course $x=lny$.
Then, helping myself out with a drawing, I concluded that the total volume of the solid must've been:
$$V=\pi\int_{0}^{1}1^2 \ dy \ +(\pi\int_{1}^{e}1^2 \ dy \ - \pi \int_{1}^{e}ln^2y \ dy)$$
However this leads me to a wrong result; I must be getting something wrong, probably in the second integration(from 1 to e)...

 

[Mark44]

Homework Statement:: Calculate the volume of the solid generated by rotating around the y-axis the plane region delimited by curves:
$y=e^x$
$x=0$
$x=1$
$y=0$
Relevant Equations:: Definite integrals definition

First, I calculated the inverse of $y=e^x$ since we're talking about y-axis rotations, which is of course $x=lny$.
Then, helping myself out with a drawing, I concluded that the total volume of the solid must've been:
$$V=\pi\int_{0}^{1}1^2 \ dy \ +(\pi\int_{1}^{e}1^2 \ dy \ - \pi \int_{1}^{e}ln^2y \ dy)$$
However this leads me to a wrong result; I must be getting something wrong, probably in the second integration(from 1 to e)...

If you integrate using shells, the integral is much simpler. With this method, the typical volume element is $\Delta V = 2\pi \text{radius} \cdot \text{height} \cdot \text{width} = 2\pi x e^x \Delta x$. The shells run from x = 0 to x = 1.

 

[hicetnunc]

To me, your equation for the volume $V$ looks fine, so if you don't get the correct solution I think you may have made a mistake later during your evaluation of the integrals. In particular, the integral $\int_{1}^{e}ln^2y \ dy$ looks a little nasty. Have you tried partial integration?

Otherwise, Mark44 gives a nice alternative way to solve for $V$. That integral might look intimidating at first, but it has the advantage that its solution can usually be found in "cheat sheets".

 

[greg_rack]

To me, your equation for the volume $V$ looks fine, so if you don't get the correct solution I think you may have made a mistake later during your evaluation of the integrals. In particular, the integral $\int_{1}^{e}ln^2y \ dy$ looks a little nasty. Have you tried partial integration?

Otherwise, Mark44 gives a nice alternative way to solve for $V$. That integral might look intimidating at first, but it has the advantage that its solution can usually be found in "cheat sheets".

Yup, I have managed to solve the integral by parts... and my procedure was actually correct; I only forgot to multiply a term by $\pi$

I gave this problem a chance by mere intuition, without having yet studied definite integrals for such geometrical applications, and I believe that's why the method I came up with looks anything but convenient!
Definitely going to go with @Mark44's :)

Thanks guys