calculus volume integration question

Question

I was wondering if anyone could help me with this volume integration problem: Find the volume of region bounded by the curve $y=e^{x^2}$ and the lines $y=0, x=0$, and $x=1$ revolving around the $y$-axis.

When I did it I got $\pi$ as the answer. Is this correct?

Answer 1

As I understand it, you are taking the curve $y=e^{x^2}$ from $x=0\text{ to }1$ and making a solid of revolution about the $y$-axis. This volume can be found with Pappus's ($2^{nd}$) Centroid Theorem, which states that the volume of a body of revolution is equal to the area times the length of the path of it centroid about the axis of revolution, let's say $V=2\pi RA$, where $R$ in this case is equal to

$$R=\frac{\int_0^1 x(e^1-y(x))dx}{\int_0^1 (e^1-y(x))dx}$$

Of course, the denominator is the area, hence we can say that

$$V=2\pi \int_0^1 x(e^1-y(x))dx=2\pi\int_0^1 x(e^1-e^{x^2})dx=\pi\int_0^1 (e^1-e^{t})dt$$

where $t=x^2$. Thus, we get the volume

$$V=\pi$$

This was verified by numerical integration. Your solution is correct.

Answer 2

Washer Method: $$\int_{0}^{e}\pi(1)^{2}dy-\int_{1}^{e}\pi(\sqrt{ln(y)})^{2}dy$$ Shell Method: $$\int_{0}^{1}2\pi x e^{x^{2}}dx$$ And the result should be $$(e-1)\pi$$