Find the volume between $z=\sqrt{x^{2}+y^{2}}$ and $x^2+y^2+z^2=2$ in spherical cordinates

Question

I am asking to find the volume of the volume trap above the cone $z=\sqrt{x^{2}+y^{2}}$ and below the sphere $x^2+y^2+z^2=2$

When I checked the solution I noticed that it was writen as $$V=\int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{2}} r^{2} \sin \theta \,d r \,d \theta \,d \varphi$$ and my question is why the boundries of $\theta$ is between $0$ to $\frac{\pi}{4}$ and not $0$ to $\pi$.

why $0$ to $\pi$ is wrong? I just can't imagine the scenerio in my head

Answer 1

as you can see the $\theta$ axis it's just going from 0 to $\frac{\pi}{4}$

We can also obtain this $\frac{\pi}{4}$ algebraically if our change of variable is:

$x =r\sin(\theta)\cos(\varphi)$

$y =r\sin(\theta)\sin(\varphi)$

$z=r\cos(\theta)$

we will obtain $z=\sqrt{x^{2}+y^{2}}\Rightarrow r\cos(\theta)=r\sin(\theta)\Rightarrow \tan(\theta)=1\Rightarrow\theta=\frac{\pi}{4}$

also can't be $2\pi$ because $\theta \in [0,\pi] $for more information: here

Answer 2

You seem to prefer, as commented, $\;\phi\;$ as azimut angle and $\;\theta\;$ as the vertical (or inclination) one. Fine. Then we have

$$\begin{cases}x=r\cos\phi\sin\theta\\{}\\ y=r\sin\phi\sin\theta\\{}\\ z=r\cos\theta\end{cases}$$

and the Jacobian is $\;r^2\sin\theta\;$ (this may be pretty confusing to physics and engineering students...).

Since the intersection of the surfaces gives

$$z^2=x^2+y^2=2-x^2-y^2\implies x^2+y^2=1$$

we get on the surfaces, projecting on the plane $\;x=0\;$ , that $$\;z^2=y^2,\,y^2+z^2=2\implies 2y^2=2\implies y=1\;(\text{ choose the positive side...})\;$$

so on that plane we get a straight triangle wito points $\;(0,0,0)\,,\,(0,1,1)\,,\,(0,0,1)\;$, and from here that the verical angle goes from $\;0\;$ radians (on the $z\,-$ axis) and until the line through $\;(0,0,0), (0,1,1)\;$, which is at $\;\pi/4\;$ radias from the $\;z\,-$ axis (or from the $\;y-$ axis, it's the same in this case), and from here that $\;0\le \theta\le \pi/4\;$, so we finally get the integral

$$\int_0^1\int_0^{2\pi}\int_0^{\pi/4} r^2\sin \theta\,d\theta\,d\phi\,dr$$

Answer 3

Not technically am answer, just an alternative solution.

We can also solve this problem using solids of rotation. Take the area bounded by $y=x$, $y=\sqrt{2-x^2}$, and the $x$ axis. Then rotate it about the $x$ axis. Our volume, then is $$V=\pi\int_0^{x_*} x^2 \mathrm{d}x + \pi\int_{x_*}^{\sqrt{2}} (\sqrt{4-x^2})^2\mathrm{d}x$$

$x_*$ is the point where the upper boundary switches from the straight line to the circular arc. Using trigonometry we can deduce $x_*=\sqrt{2}\cos(\pi/4)=1$. So our volume is $$V=\pi\left(\int_0^1 x^2 \mathrm{d}x +\int_1^{\sqrt{2}} 4-x^2 \mathrm{d}x\right)=\frac{10\pi}{3}(\sqrt{2}-1)\approx 1.183.$$

Answer 4

Intersection of $x^2+y^2+z^2=2$ and $z=\sqrt{x^{2}+y^{2}}$ on $OXY$ plane gives $x^2+y^2=1$ so on Cartesian coordinates we have $$\int\limits_{-1}^{1}\int\limits_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int\limits_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^2-y^2}}dxdydz = 4\int\limits_{0}^{1}\int\limits_{0}^{\sqrt{1-x^2}}\int\limits_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^2-y^2}}dxdydz$$ Now let's take spherical coordinates: \begin{array}{} x = r \sin \phi \cos \theta; \\ y = r \sin \phi \sin \theta; \\ z = r \cos \phi \end{array}

From $z$ coordinate bounds $$r \sin \phi \leqslant r \cos \phi \leqslant \sqrt{2-r^2 \sin^2 \phi}$$

Left inequality gives $\sin \phi \leqslant \cos \phi$, from which we can obtain $\phi \leqslant \frac{\pi}{4}$. Right inequality gives $r \leqslant \sqrt{2}$. So for integral in spherical coordinates we have $$4\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{\pi}{4}}\int\limits_{0}^{\sqrt{2}} r^{2} \sin \theta d r d \phi d \theta$$