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:

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

$\endgroup$
3
  • $\begingroup$ Why don't you use cylindrical coordinates? $\endgroup$
    – Andrei
    Commented Oct 28, 2020 at 13:49
  • 1
    $\begingroup$ You are doing an integral of $z dV$ instead of an integral of $dV$ $\endgroup$
    – Andrei
    Commented Oct 28, 2020 at 13:51
  • 1
    $\begingroup$ Also, the integration limits are wrong $\endgroup$
    – Andrei
    Commented Oct 28, 2020 at 13:53

3 Answers 3

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 -

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

$\endgroup$
3
  • $\begingroup$ 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: $\int \int \int _{R} {\sqrt {4-r^2}rdzdrd\theta }$ $\endgroup$ Commented Oct 28, 2020 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
    Commented Oct 28, 2020 at 18:58
  • $\begingroup$ Thank you very much! This is the internet at its best. $\endgroup$ Commented Oct 28, 2020 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 $-\pi/4$

$\endgroup$
2
  • 2
    $\begingroup$ please note that $y \geq 0$ so $ -\pi/4 \leq \theta \leq 0$ should not be considered. $\endgroup$
    – Math Lover
    Commented Oct 28, 2020 at 15:11
  • 1
    $\begingroup$ Sorry @MathLover, I've missed that. You are right $\endgroup$
    – Andrei
    Commented Oct 28, 2020 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.

$\endgroup$
6
  • 2
    $\begingroup$ How will this bind the area in XY plane to $x^2 + y^2 \leq 1$? $\endgroup$
    – Math Lover
    Commented Oct 28, 2020 at 15:09
  • 1
    $\begingroup$ My mistake, I missed that bit of information. Fixed $\endgroup$
    – user170231
    Commented Oct 28, 2020 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
    Commented Oct 28, 2020 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
    Commented Oct 28, 2020 at 17:51
  • $\begingroup$ Thank you for pointing that out. I've updated my answer with the missing chunk of $K$. $\endgroup$
    – user170231
    Commented Oct 28, 2020 at 18:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .