Volume of region bounded by $x^2+y^2 \leq 4$, $x \geq 0$, $0<z<x^2-y^2$
I have used the volume formula but not sure of the limits of integration.If I take the limits like the below, I get the answer zero.
$V = \int \int \int dzdydx$
$V =\int_{0}^{2} \int_{0}^{4-x^2} \int_{0}^{x^2-y^2} dzdydx$
Is the Limit setting correct? Please suggest an easy way to solve this.
On
After Felix Marin's prompt, I realized there was a mistake in my upper bound of z which led to wrong result. I had written it only in terms of $y$ which is not true. It is a function of both $x$ and $y$. So the best way to proceed is using cylindrical coordinates.
We have been given below bounds for the region -
i) $x^2 + y^2 \le 4$ or $x \le r \cos \theta, y \le r \sin \theta$ where $r = 2$.
ii) $x \ge 0 \,$ i.e $ \,r \cos\theta \ge 0$ or $\cos \theta \ge 0 \implies (-\frac{\pi}{2} \le \theta \le \frac{\pi}{2})$.
iii) $0 \le z \le x^2 - y^2 = r^2(cos^2 \theta - \sin^2 \theta) = r^2 \cos 2\theta $.
As $z \ge 0 \implies$ $\, cos^2\theta \ge sin^2\theta$ or $\, (-\frac{\pi}{4} \le \theta \le \frac{\pi}{4})$ [based on restriction given by (ii)]
Now the integration -
$ \begin{align} V &= \displaystyle\int_{-\pi/4}^{\pi/4} \int_0^2 \int_{0}^{r^2 \cos2\theta} dz \, r \,dr \,d\theta \\ \\ &= \displaystyle \int_{-\pi/4}^{\pi/4} \int_0^2 r^3 \, \cos2\theta \, \,dr \,d\theta \\ \\ &= \displaystyle \int_{-\pi/4}^{\pi/4} [\frac {r^4}{4}]_0^2 \, \cos2\theta \, \,d\theta \\ \\ &= \displaystyle 4 \int_{-\pi/4}^{\pi/4} \cos2\theta \, d\theta \\ \\ &= \displaystyle 2 [\sin2\theta]_{-\pi/4}^{\pi/4} = 4\\ \end{align} $
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} V & \equiv \bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}} \bracks{x^{2} + y^{2} \leq 4}\bracks{x \geq 0} \bracks{0 < z < x^{2} - y^{2}}\dd x\,\dd y\,\dd z} \\[5mm] = &\ \underbrace{\int_{0}^{2\pi}\int_{0}^{\infty}\int_{0}^{2} \bracks{\rho\cos\pars{\phi} \geq 0} \bracks{z < \rho^{2}\cos\pars{2\phi}}\rho\,\dd\rho\,\dd z\,\dd\phi} _{\ds{Cylindrical\ Coordinates}} \\[5mm] = &\ \int_{-\pi}^{\pi}\int_{0}^{\infty}\int_{0}^{2} \bracks{-\cos\pars{\phi} \geq 0} \bracks{z < \rho^{2}\cos\pars{2\phi}}\rho\,\dd\rho\,\dd z\,\dd\phi \\[5mm] = &\ 2\int_{0}^{\pi}\int_{0}^{\infty}\int_{0}^{2} \bracks{-\cos\pars{\phi} \geq 0} \bracks{z < \rho^{2}\cos\pars{2\phi}}\rho\,\dd\rho\,\dd z\,\dd\phi \\[5mm] = &\ 2\int_{-\pi/2}^{\pi/2}\int_{0}^{\infty}\int_{0}^{2} \bracks{\sin\pars{\phi} \geq 0} \bracks{z < -\rho^{2}\cos\pars{2\phi}}\rho\,\dd\rho\,\dd z\,\dd\phi \\[5mm] = &\ 2\int_{0}^{\pi/2}\int_{0}^{\infty}\int_{0}^{2} \bracks{z < -\rho^{2}\cos\pars{2\phi}}\rho\,\dd\rho\,\dd z\,\dd\phi \\[5mm] = &\ \int_{0}^{\pi}\int_{0}^{\infty}\int_{0}^{2} \bracks{z < -\rho^{2}\cos\pars{\phi}}\rho\,\dd\rho\,\dd z\,\dd\phi \\[5mm] = &\ \int_{\pi/2}^{\pi}\int_{0}^{\infty}\int_{0}^{2} \bracks{z < -\rho^{2}\cos\pars{\phi}}\rho\,\dd\rho\,\dd z\,\dd\phi \\[5mm] = &\ {1 \over 2}\int_{0}^{\pi/2}\int_{0}^{4}\int_{0}^{\infty} \bracks{z < \rho\sin\pars{\phi}}\,\dd z \,\dd\rho\,\dd\phi \\[5mm] = &\ {1 \over 2}\int_{0}^{\pi/2}\int_{0}^{4}\rho\sin\pars{\phi} \,\dd\rho\,\dd\phi = \bbx{\large\color{red}{4}} \\ & \end{align}