Finding the volume between $y = e^x$ and $y = 0$ and $x = 0$ and $x = 1$ about the x-axis

Question

I'm trying to do this problem using shells. Using cross-sections, I know that answer should be ${\pi \over 2}[e^2 - 1]$. However, this is not the answer I get.

So I know that the radius of a typical cylinder for this function will be $y$ and the height will be $1 - \ln(y)$. So, I get

$2\pi\int_{1}^ey(1 - \log(y)) = 2\pi[{y^2\over2} - {y^2\log(y)\over2} + {y^2\over4}]_{1}^{e} = {\pi\over2}[e^2 - 3]$. I'm not sure what I'm doing wrong.

Thanks.

Answer 1

The height is not always $1- \ln (y)$, that only holds true for $y \ge 1$.

If you add the volume of the cylinder at the bottom which correspond to when the value of $y$ is between $0$ and $1$ which is $\pi$

$$\frac{\pi}{2}[e^2-3]+\pi= \frac{\pi}{2}[e^2-1]$$

Answer 2

Hint:

$$\text{Volume}=\pi\int_0^1y^2dx=\pi\int_0^1e^{2x}dx$$