# Volume with spherical polar coordinates

Determine the volume between the surface $$z=\sqrt{4-x^2-y^2}$$ and the area of the xy plane determined by $$x^2+y^2\le 1,\ x+y>0,\ y\ge 0$$.

I convert to spherical polar coordinates.

$$K=0\le r\le 1,\ 0\le \phi \le \frac{3\pi}{4},\ 0\le \theta \le 2\pi$$

$$\iiint_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi drd\phi d\theta$$

I can't figure out how to take $$\int_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi dr$$, which makes me think I made a mistake somewhere.

EDIT: Thanks for all the answers.

Now I understand how the limits of $$\theta ,r,z$$ works.

I don't fully understand where the function "disappear".

$$\sqrt {4-x^2-y^2} =\sqrt {4-r^2}$$

Why isn't it then:

$$\int \int \int _{K} {\sqrt {4-r^2}rdzdrd\theta }$$

• Why don't you use cylindrical coordinates? – Andrei Oct 28 at 13:49
• You are doing an integral of $z dV$ instead of an integral of $dV$ – Andrei Oct 28 at 13:51
• Also, the integration limits are wrong – Andrei Oct 28 at 13:53

Area on XY plane is bound by $$x^2 + y^2 \leq 1, y \geq 0, x + y \geq 0$$

This is a sector of the circle $$x^2 + y^2 \leq 1$$ bound between positive $$X$$-axis and line $$y = -x$$ in the second quadrant. This comes from the fact that $$y \geq 0$$ so part of the circle in third and fourth quadrant of $$XY$$ plane is not included. $$x + y \geq 0$$ is true for quarter of the circle in the first quadrant as both $$x$$ and $$y$$ are positive. It is also true for part of the circle in the second quadrant above line $$y = -x$$ as $$|y| \geq |x|$$.

Now you are asked to find the volume between this area on XY plane and $$z = \sqrt{4-x^2-y^2}$$. So it is essentially a cylinder ($$\frac{3}{8}$$ cross section of a cylinder of radius $$1$$) cut out of the sphere of radius $$2$$ above $$XY$$ plane.

So here is how it will look in cylindrical coordinates -

$$\displaystyle \int_{0}^{3\pi/4} \int_{0}^{1} \int_{0}^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta$$

• Thanks for your answer. Now I understand how the limits of $\theta ,r,z$ works. I don't fully understand where the function "disappear". $\sqrt {4-x^2-y^2} =\sqrt {4-r^2}$ Why isn't it then: $\int \int \int _{R} {\sqrt {4-r^2}rdzdrd\theta }$ – user9060784 Oct 28 at 18:44
• @user9060784 $\int dV = \iiint_R r dz dr d \theta$. Please note if I take a tiny block of thickness $dz$ along $z$ axis, the volume will be given by area of it in $XY$ plane (or parallel to it) multiplied by thickness $dz$. The area is $(r d\theta) \times dr$, where $r d\theta$ is the length of the block and $dr$ its width. So the volume $dV = r d\theta \, dr \, dz$. This is what you integrate to get the volume. – Math Lover Oct 28 at 18:58
• Thank you very much! This is the internet at its best. – user9060784 Oct 28 at 20:17

This is much easier to solve in cylindrical coordinates. $$x=r\cos\theta\\y=r\sin\theta\\z=h$$ Then the limits for $$r$$ are $$0$$ and $$1$$, the limits for $$\theta$$ are from $$-\frac\pi4$$ to $$\frac{3\pi}4$$, and the limits for $$h$$ are $$0$$ and $$4-r^2$$. With these, $$V=\int_{-\frac\pi4}^{\frac{3\pi}4}d\theta\int_0^1dr\cdot r\int_0^{\sqrt{4-r^2}}dh$$

Note see comment below. Since $$y>0$$, the lower limit for $$\theta$$ is $$0$$, not $$-\pi/4$$

• please note that $y \geq 0$ so $-\pi/4 \leq \theta \leq 0$ should not be considered. – Math Lover Oct 28 at 15:11
• Sorry @MathLover, I've missed that. You are right – Andrei Oct 28 at 15:44

Using spherical coordinates, you would have to split up $$K$$ into two regions,

$$K_1=\left\{(r,\theta,\phi)\mid 0\le r\le2,0\le\theta\le\frac{3\pi}4,0\le\phi\le\frac\pi6\right\}$$

$$K_2=\left\{(r,\theta,\phi)\mid0\le r\le\sqrt{\csc\phi},0\le\theta\le\frac{3\pi}4,\frac\pi6\le\phi\le\frac\pi2\right\}$$

(where $$x=r\cos\theta\sin\phi$$, $$y=r\sin\theta\sin\phi$$, and $$z=r\cos\phi$$). The upper limit on $$\phi$$ for $$K_1$$ and lower limit for $$K_2$$ come from the intersection of the cylinder $$x^2+y^2=1$$ and the sphere $$z=\sqrt{4-x^2-y^2}$$. On the sphere, $$r=2$$, so we have

$$2\cos\phi=\sqrt3\implies\phi=\cos^{-1}\left(\frac{\sqrt3}2\right)=\frac\pi6$$

The upper limit for $$r$$ in $$K_2$$ is obtained by converting the equation of the cylinder $$x^2+y^2=1$$ into spherical coordinates:

$$(r\cos\theta\sin\phi)^2+(r\sin\theta\sin\phi)^2=r^2\sin^2\phi=1\implies r=|\csc\phi|=\csc\phi$$

Then the volume is

$$\int_0^{\frac\pi6}\int_0^{\frac{3\pi}4}\int_0^2r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi+\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\int_0^{\csc\phi}r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi$$

The first integral is trivial. For the second, integrating with respect to $$r$$ yields

$$\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\int_0^{\csc\phi}r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi=\frac13\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\csc^2\phi\,\mathrm d\theta\,\mathrm d\phi$$

and observing that $$\csc^2\phi=\frac{\mathrm d}{\mathrm d\phi}(-\cot\phi)$$, it turns out the second integral is, too.

• How will this bind the area in XY plane to $x^2 + y^2 \leq 1$? – Math Lover Oct 28 at 15:09
• My mistake, I missed that bit of information. Fixed – user170231 Oct 28 at 16:31
• I think this is still wrong. For example, point $(1,0,0)$ is on the edge of this volume. This means $r\ne 0$ and $\cos\phi=0$, or $\phi=\frac\pi 2$. Since you cover the entire volume above the xy plane, $\phi$ will be between $0$ and $\pi/2$, but the limit on the radius is a function of $\phi$. – Andrei Oct 28 at 17:48
• @Andrei agree. This will give the volume of the spherical cone, not of the whole cylindrical region. It will be somewhat complicated in spherical coordinates, best to use cylindrical coordinates. – Math Lover Oct 28 at 17:51
• Thank you for pointing that out. I've updated my answer with the missing chunk of $K$. – user170231 Oct 28 at 18:57