Recent content by notorious_lx

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...

r would go from 0 to 1. so the equation would be \int_0^{2\pi} \int_0^1 (e^{-r^2})rdrd\theta ?

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 boundary 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...