Calculating volume of solid

Question

Let $$M_1 = \{(x,y,z) \in \Bbb R^3 : (x,y) \in D,\ 0 \leq z \leq yx2 + y^3\}$$ where $$D =\{(x, y) \in\Bbb R^2 : |x| \leq y, 1 \leq x^2 + y^2 \leq 4\}$$ Calculate the volume, $V(M_1)$, of $M_1$.

My work:

Answer 1

You haven't converted the domain $D$ into boundaries yet. The condition $|x|\le y$ is equivalent to $-y\le x\le y$ and requires $y\ge0$. If $x=y$ and $y\ge0$, then in polar coordinates $\tan\theta=1$ and $\theta=\frac{\pi}4$. Similarly, if $x=-y$ and $y\ge0$, $\tan\theta=-1$ and $\theta=\frac{3\pi}4$.

Also $1\le x^2+y^2\le4$ translates in polar coordinates to $1\le r^2\le4$, or $1\le r\le2$.

Then we can write $yx^2+y^3=(x^2+y^2)y=r^2\cdot r\sin\theta$ in polar coordinates, so your integral for volume in cylindrical coordinates becomes $$\begin{align}V(M_1) & =\int_{\frac{\pi}4}^{\frac{3\pi}4}\int_1^2\int_0^{r^3\sin\theta}dz\,r\,dr\,d\theta \\ & =\int_{\frac{\pi}4}^{\frac{3\pi}4}\int_1^2r^4\sin\theta\,dr\,d\theta \\ & =\int_{\frac{\pi}4}^{\frac{3\pi}4}\frac{31}5\sin\theta\,d\theta \\ & =\frac{31}5\sqrt2\end{align}$$