This is correct.
There is a cylinder with height 4. When using a double integral to find the volume of a solid object, you can set it up with the "Top - Bottom" as the function to integrate. This can also be done by adding in a third integral and integrating 1.
\int_0^{2\pi} \int_0^3 \int_0^4...
ok i had \int_0^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (e^{-(x^2+y^2)})
Then I did this and used r^2=1, but this is only on the boundry so this is incorrect.
\begin{align}
&= \sqrt{1-x^2}\\
&= \sqrt{r^2-r^2\cos^2{\theta}}\\
&= r\sqrt{1-\cos^2{\theta}}\\
&= r\sin{\theta}
\end{align}
OK here is what I have so far..
\int_0^{2\pi} \int_{-rsinθ}^{rsinθ} (e^{-r^2}-{\frac{1}{e}})rdrdθ + \int_0^{2\pi} \int_{-rsinθ}^{rsinθ} ({\frac{1}{e}}rdrdθ)
I'm not sure if this is right.. I may be overthinking this way too much.. haha
edit: This is just equal to \int_0^{2\pi}...
1. Ok, so the question is.. Find the exact volume of the solid bounded above by the surface z=e^{-x^2-y^2}, below by the xy-plane, and on the side by x^2+y^2=1.
2. Alright. So, I know that I can use a double integral to find the volume, and switching to polar coordinates would be simpler...