3
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ Why don't you use cylindrical coordinates? $\endgroup$ – Andrei Oct 28 at 13:49
  • 1
    $\begingroup$ You are doing an integral of $z dV$ instead of an integral of $dV$ $\endgroup$ – Andrei Oct 28 at 13:51
  • 1
    $\begingroup$ Also, the integration limits are wrong $\endgroup$ – Andrei Oct 28 at 13:53
2
$\begingroup$

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 -

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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: $\int \int \int _{R} {\sqrt {4-r^2}rdzdrd\theta }$ $\endgroup$ – user9060784 Oct 28 at 18:44
  • 1
    $\begingroup$ @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. $\endgroup$ – Math Lover Oct 28 at 18:58
  • $\begingroup$ Thank you very much! This is the internet at its best. $\endgroup$ – user9060784 Oct 28 at 20:17
2
$\begingroup$

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 $-\pi/4$

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ please note that $y \geq 0$ so $ -\pi/4 \leq \theta \leq 0$ should not be considered. $\endgroup$ – Math Lover Oct 28 at 15:11
  • 1
    $\begingroup$ Sorry @MathLover, I've missed that. You are right $\endgroup$ – Andrei Oct 28 at 15:44
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ How will this bind the area in XY plane to $x^2 + y^2 \leq 1$? $\endgroup$ – Math Lover Oct 28 at 15:09
  • 1
    $\begingroup$ My mistake, I missed that bit of information. Fixed $\endgroup$ – user170231 Oct 28 at 16:31
  • 3
    $\begingroup$ 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 $\phi$. $\endgroup$ – Andrei Oct 28 at 17:48
  • 3
    $\begingroup$ @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. $\endgroup$ – Math Lover Oct 28 at 17:51
  • $\begingroup$ Thank you for pointing that out. I've updated my answer with the missing chunk of $K$. $\endgroup$ – user170231 Oct 28 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.