Integrate $f(x,y)=e^{x^{2}+y^{2}}$ over the region $1 \leq x^{2}+y^{2} \leq 2$.
Solution:
\begin{align*}
&\iint_{1 \leq x^{2}+y^{2} \leq 2} e^{x^{2}+y^{2}} dxdy =
\int_{0}^{2\pi} \int_{1}^{\sqrt{2}}e^{r^{2} \cos^{2}\theta + r^{2} \sin^{2}\theta}rdrd\theta=\\
&=\int_{0}^{2\pi} \int_{1}^{\sqrt{2}} e^{r^{2}}rdrd\theta=2 \pi \int_{1}^{\sqrt{2}}e^{r^{2}}rdr=
\pi \left. e^{r^{2}}\right|_{1}^{\sqrt{2}} = \pi (e^{2}-e).
\end{align*}
Use the triple integral to find the volume of the following tetrahedron with the vertices
$(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$.
Solution:
First of all you need to draw a picture. If we recall how I explained you to find the limits of such triple integrals then you can come up to the following
\begin{align*}
&\text{Volume} = \int_{0}^{1} \int_{0}^{1-y} \int_{0}^{1-y-x}dzdxdy=
\int_{0}^{1} \int_{0}^{1-y} 1-y-x dxdy = \int_{0}^{1}(1-y)^{2} - \frac{(1-y)^{2}}{2}dy=\\
&\int_{0}^{1} \frac{(1-y)^{2}}{2} = -\left. \frac{(1-y)^{3}}{6}\right|_{0}^{1} = \frac{1}{6}
\end{align*}