Evaluate the volume of $V\subset \mathbb{R}^3$, which is bounded by paraboloid $z=1-x^2-y^2$ and the surface $z=1-y,$ for $z\geqslant 0.$
Attempt. The desired volume goes like:
$$\iint_D \big(1-x^2-y^2-(1-y)\big)\,dx\,dy,$$
where $D$ is the projection of V on $\mathbb{R}^2$. How do we determine $D$?
Thanks in advance.
Best Answer
I think that it is more natural to use cilindrical coordinates. Then\begin{align}1-y\leqslant1-x^2-y^2&\iff1-r\sin\theta\leqslant1-r^2\\&\iff r\sin\theta\geqslant r^2\\&\iff\sin\theta\geqslant r.\end{align}So, compute the integral$$\int_0^\pi\int_0^{\sin\theta}\int_{1-r\sin\theta}^{1-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$You should get $\dfrac\pi{32}$.