Find the volume inside the paraboloid $az=x^2+y^2$ and inside the sphere $x^2+y^2+z^2=2a^2$
I asked this question before here , but now I am trying a different way to solve it and I cannot figure out why it is not working.
What I did:
I checked which of them is above that other , and figured it is the sphere , then I found the intersection point between them which is $2a^2-z^2=az$ , $z=a$.
I tried to solve with sphere coordinates , so I transformed the equations to sphere coordinates and got $arcos\phi=r^2sin^2\phi cos^2\theta+r^2sin^2\phi sin^2\theta$ , $acos\phi=rsin^2\phi$ (this is for the paraboloid) for the sphere I got $r=\sqrt{2}a$ Then I compared them and got $cos\phi=\sqrt{2}sin^2\phi$ So I need to find $\phi$ at the intersection points what I did was $sin^2\phi=1-cos^2\phi$ , $cos\phi=\sqrt{2}(1-cos^2\phi)$ , $cos\phi-\sqrt{2}+\sqrt{2}cos^2\phi=0$ then after doing $cos\phi=t$ I got $\phi=\frac{\pi}{4} $ and $\phi=\frac{7\pi}{4} $ I only took $\phi=\frac{\pi}{4}$ because as far as I know in sphere coordinates $0\leq\phi\leq\pi$.
After that I have 2 volumes , the first until the paraboloid and the second till the sphere. so for paraboloid I got $V_P $ \begin{cases} 0\leq\theta\leq2\pi\\ \frac{\pi}{4}\leq\phi\leq\frac{\pi}{2}\\ 0\leq r\leq a\\ \end{cases} And for sphere $V_S$ \begin{cases} 0\leq\theta\leq2\pi\\ 0\leq\phi\leq\frac{\pi}{4}\\ 0\leq r\leq \sqrt2a\\ \end{cases}
After that the integration result is wrong , here is what I did $$\int_0^{2\pi}\int_{\pi/4}^{\pi/2}\int_0^{a}r^2\cdot sin\phi\,\mathrm dr\,\mathrm d\phi\,\mathrm d\theta+\int_0^{2\pi}\int_0^{\pi/4}\int_0^{\sqrt{2}a}r^2\cdot sin\phi\,\mathrm dr\,\mathrm d\phi\,\mathrm d\theta$$
according to my previous post the answer should be $\left(\frac{ \left(8 \sqrt{2}-7\right) \pi a^3}6\right).$ while what I got is $\frac{\sqrt{2}\pi a^3}{3}$ + $\frac{4\sqrt{2}\pi a^3-4\pi a^3}{3}$
I am trying to rely on this way because I noticed that my teacher does this a lot , but I am not even sure if it works here? or if what I did is right in someway.
I get the same result with your integral as you write. (Although it would be easier to compare with the referenced answer in the form $$ \frac{(5 \sqrt{2} - 4)\pi a^3}{3} \text{.)} $$
Your integral with $\phi \in [\pi/4, \pi/2]$ has $r \in [0,a]$, suggesting that the lower part of the volume is a ball of radius $a$. That is, there is a discontinuous jump in the radius at $\phi = \pi/4$ and the radius does not decrease as $\phi$ goes to $\pi/2$. This is incorrect, the lower part of the volume is bounded by the paraboloid. Instead, the upper bound for $r$ should smoothly decrease from $\sqrt{2} a$ to $0$.
Consider the slice where $y = 0$. On this slice, we have the parabola $a z = x^2$, so $\frac{z}{x} = \frac{x}{a}$. Of course, $\tan (\pi/2 - \phi) = \frac{z}{x}$, so you should be able to get the upper bound on $r$ as a function of $\phi$, easily enough...
(Added in response to a comment...)
There are a few ways to get the upper bound on $r$ for the integral with $\phi \in [\pi/4, \pi/2]$. But first, let's get a picture so we can see what we are doing. (Step one in "how to solve it": sketch a picture.) (This picture is shown with $a = 4/3$, but one obtains essentially the same picture, with different (identical) ranges on the three axes, for different choices of $a$.)
This helps us see that the sphere provides the bound on the radius for $\phi$ small and the paraboloid provides the bound on the radius for $\phi$ large. This is a volume of revolution, so we can take the slice with $y = 0, x \geq 0, z \geq 0$ (equivalently, $\theta = 0, 0 \leq \phi \leq \pi/2$).
Let's introduce $\rho = \sqrt{x^2 + y^2}$, the distance between the projection of a point onto the $xy$-plane parallel to the $z$-axis and the origin (of any coordinate system in use, since the origins are all the same point). Then your $3$-dimensional radial coordinate satisfies $r^2 = \rho^2 + z^2$. Since this is a volume of revolution, we get the same diagram for every choice of $\theta$.
The points on the parabola satisfy $$ a z = \rho^2 \text{.} $$ One way forward is to divide both sides by $\rho$, obtaining $$ a \frac{z}{\rho} = \rho $$ so that $$ a \tan\left( \frac{\pi}{2} - \phi \right) = \rho \text{.} $$ and when continue manipulating to make $r$ appear on the right-hand side.
A different way to proceed is to start by immediately making $r$ appear on the right-hand side: $$ a z + z^2 = \rho^2 + z^2 = r^2 \text{.} $$ So now the upper bound on $r$ is $\sqrt{az + z^2}$. Unfortunately, $z$ is not a variable available to us in the innermost integral. But $z = r \cos \phi$, so ... \begin{align*} r^2 &= a z + z^2 \\ &= ar \cos \phi + r^2 \cos^2 \phi \text{,} \end{align*} so \begin{align*} r^2(1 - \cos^2 \phi) - a r \cos \phi &= 0 \text{, and } \\ r(r(1 - \cos^2 \phi) - a \cos \phi) &= 0 \text{.} \end{align*}
The two roots are $r = 0$ and $$ r = \frac{a \cos \phi}{1 - \cos^2 \phi} [{} = a \cot \phi \csc \phi ] \text{.} $$ The root $r = 0$ tells us our ray intersects the paraboloid at $r = 0$, giving the lower bound for the $r$ integral. The upper bound is $a \cos \phi / (1 - \cos^2 \phi)$.
And with this change, replacing the $\phi \in [\pi/4, \pi/2]$ integral with $$ \int_0^{2\pi} \; \int_{\pi/4}^{\pi/2} \; \int_0^{\frac{a \cos \phi}{1 - \cos^2 \phi}} \; r^2 \,\mathrm{d}r \,\mathrm{d}\phi \,\mathrm{d}\theta $$ produces the same result as does the previous answer you cite.