# Calculate the volume of a part of a sphere inside a cylinder

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• Jun 13th 2009, 05:05 PM
arbolis
Calculate the volume of a part of a sphere inside a cylinder
Find the volume of the part of the sphere $r^2+z^2=a^2$ which is inside the cylinder $r=a \sin (\theta)$ where $(r,z,\theta)$ are the cylindrical coordinates.

My attempt : I'm not able to visualize the sphere nor the cylinder.
For the sphere, does $r^2=x^2+y^2$? If so then I can imagine the sphere. (Centered at the origin $(0,0,0)$ and with radius $a$).
For the cylinder $\sqrt{x^2+y^2}=a \sin (\theta)$ I don't know how to go further. I also feel it's useless to convert all in Cartesian coordinates but this way it's easier to me to visualize the volumes.
• Jun 13th 2009, 05:46 PM
mr fantastic
Quote:

Originally Posted by arbolis
Find the volume of the part of the sphere $r^2+z^2=a^2$ which is inside the cylinder $r=a \sin (\theta)$ where $(r,z,\theta)$ are the cylindrical coordinates.

My attempt : I'm not able to visualize the sphere nor the cylinder.
For the sphere, does $r^2=x^2+y^2$? If so then I can imagine the sphere. (Centered at the origin $(0,0,0)$ and with radius $a$).
For the cylinder $\sqrt{x^2+y^2}=a \sin (\theta)$ I don't know how to go further. I also feel it's useless to convert all in Cartesian coordinates but this way it's easier to me to visualize the volumes.

$r = a \sin \theta \Rightarrow r = a \frac{y}{r} \Rightarrow r^2 = a y \Rightarrow x^2 + y^2 - ay = 0$.

Now complete the square etc.
• Jun 13th 2009, 06:51 PM
arbolis
So the sphere is centered at the origin and has a radius of $a$.
The cylinder is centered in $(0,\frac{a}{2})$ and has a radius of $\frac{a}{4}$ and is parallel to the z-axis.
I now can visualize the volume. However I don't know how to set up the triple integrals. Should I use Cartesian coordinates? Must I find the ellipses of intersection between the sphere and the cylinder? How would you proceed?
• Jun 14th 2009, 02:25 PM
Jhevon
Quote:

Originally Posted by arbolis
Find the volume of the part of the sphere $r^2+z^2=a^2$ which is inside the cylinder $r=a \sin (\theta)$ where $(r,z,\theta)$ are the cylindrical coordinates.

My attempt : I'm not able to visualize the sphere nor the cylinder.
For the sphere, does $r^2=x^2+y^2$? If so then I can imagine the sphere. (Centered at the origin $(0,0,0)$ and with radius $a$).
For the cylinder $\sqrt{x^2+y^2}=a \sin (\theta)$ I don't know how to go further. I also feel it's useless to convert all in Cartesian coordinates but this way it's easier to me to visualize the volumes.

$r^2 + z^2 = a^2$ is simply a sphere of radius $a$ centered at the origin.

following Mr F's lead, the cylinder is $x^2 + \left( y - \frac a2 \right)^2 = \left( \frac a2 \right)^2$

in the xy-plane, the trace of this cylinder is a circle with radius a/2 centered at (0, a/2)

now, you are bounded above and below by the sphere. we can simply find the volume in the first quadrant, and then multiply the answer by 4.

hence, we integrate the function 1, with the following limits:

$0 \le x \le \sqrt{ay - y^2}$, $0 \le y \le a$ and $0 \le z \le \sqrt{a^2 - x^2 - y^2}$

and don't forget to multiply by 4

any questions?
• Jun 14th 2009, 02:48 PM
arbolis
Quote:

Originally Posted by Jhevon
$r^2 + z^2 = a^2$ is simply a sphere of radius $a$ centered at the origin.

following Mr F's lead, the cylinder is $x^2 + \left( y - \frac a2 \right)^2 = \left( \frac a2 \right)^2$

in the xy-plane, the trace of this cylinder is a circle with radius a/2 centered at (0, a/2)

now, you are bounded above and below by the sphere. we can simply find the volume in the first quadrant, and then multiply the answer by 4.

hence, we integrate the function 1, with the following limits:

$0 \le x \le \sqrt{ay - y^2}$, $0 \le y \le a$ and $0 \le z \le \sqrt{a^2 - x^2 - y^2}$

and don't forget to multiply by 4

any questions?

That makes perfect sense, thank you very much.
I'm having difficulties with $\int_0^{a} \sqrt{a^2-x^2-y^2} dy$ but I don't give up.
And I'm having some troubles to realize that $x$ goes from $0$ to $\sqrt{ay-y^2}$ and $z$ from $0$ to $\sqrt{a^2-x^2-y^2}$. But I'll try to derive it. If I have time I'll try to solve the problem using cylindrical coordinates.
• Jun 14th 2009, 02:54 PM
Jhevon
Quote:

Originally Posted by arbolis
That makes perfect sense, thank you very much.
I'm having difficulties with $\int_0^{a} \sqrt{a^2-x^2-y^2} dy$ but I don't give up.

that's not the integral you want to do first. you leave the one with the constant limits for the last. anyway, an integral like that would conventionally be done using trig substitution

Quote:

And I'm having some troubles to realize that $x$ goes from $0$ to $\sqrt{ay-y^2}$
in the equation for the cylinder, solve for x

Quote:

and $z$ from $0$ to $\sqrt{a^2-x^2-y^2}$
solve for z in the equation of the sphere.

then note that you only care about the first quadrant.
• Jun 14th 2009, 03:05 PM
arbolis
Quote:

Originally Posted by Jhevon
that's not the integral you want to do first. you leave the one with the constant limits for the last. anyway, an integral like that would conventionally be done using trig substitution

Yes I know, it was the second integral, the first being the one with dz.

Quote:

in the equation for the cylinder, solve for x

solve for z in the equation of the sphere.
ok thanks. I think I won't forget this in future.

Quote:

then note that you only care about the first quadrant.
yes I realized thanks to your explanations.

In cylindrical coordinates, I have a doubt about the second integral (dr) : $V=4 \int_0^{\frac{\pi}{2}} \int_0^{a} \int_0^{\sqrt{a^2-r^2}} dzdr d\theta$. (Or is it $V=4 \int_0^{\frac{\pi}{2}} \int_0^{\text{Another function}} \int_0^{\sqrt{a^2-r^2}} r dzdr d\theta$. ?)
• Jun 14th 2009, 03:10 PM
Jhevon
Quote:

Originally Posted by arbolis
Yes I know, it was the second integral, the first being the one with dz.

ok thanks. I think I won't forget this in future.

yes I realized thanks to your explanations.

In cylindrical coordinates, I have a doubt about the second integral (dr) : $V=4 \int_0^{\frac{\pi}{2}} \int_0^{a} \int_0^{\sqrt{a^2-r^2}} dzdr d\theta$. (Or is it $V=4 \int_0^{\frac{\pi}{2}} \int_0^{\text{Another function}} \int_0^{\sqrt{a^2-r^2}} r dzdr d\theta$. ?)

it is the latter. how would you find the limits for r?

it is harder here since the circle in the xy-plane is not centered at the origin
• Jun 14th 2009, 03:30 PM
the_doc
In cylindrical coordinates the integral you want needs $a \sin \theta$ in place of the "another function".
• Jun 14th 2009, 03:37 PM
Jhevon
Quote:

Originally Posted by the_doc
In cylindrical coordinates the integral you want needs $a \sin \theta$ in place of the "another function".

i wanted to see if he would have gotten that himself :p
• Jun 14th 2009, 03:42 PM
the_doc
Quote:

Originally Posted by Jhevon
i wanted to see if he would have gotten that himself :p

LOL. Fair enough --- I'm off to bed. :)
• Jun 14th 2009, 03:43 PM
arbolis
Quote:

Originally Posted by Jhevon
it is the latter. how would you find the limits for r?

it is harder here since the circle in the xy-plane is not centered at the origin

I was thinking about a $\sqrt{ \text{something}^2-(x\text{another thing} )^2}$.

Quote:

Originally Posted by the_doc
In cylindrical coordinates the integral you want needs $a \sin \theta$ in place of the "another function".

Thanks, but I don't understand why. Could you explain why is it $\frac{ay}{r}=a\sin \theta$.
• Jun 14th 2009, 03:52 PM
Jhevon
Quote:

Originally Posted by arbolis
I was thinking about a $\sqrt{ \text{something}^2-(x\text{another thing} )^2}$.

Thanks, but I don't understand why. Could you explain why is it $\frac{ay}{r}=a\sin \theta$.

$r = a \sin \theta$ was given :p
• Jun 14th 2009, 04:00 PM
arbolis
Quote:

Originally Posted by Jhevon
$r = a \sin \theta$ was given :p

lol, I can't believe I missed that. That's the change of coordinates. Believe it or not, I realized this a few minutes ago while cooking. (Nod)

Edit: I'm stuck on the last integral : $V=-\frac{4}{3} a^{\frac{3}{2}} \int _0^{\frac{\pi}{2}} \sin ^{\frac{3}{2}} (\theta) d\theta$.
• Jun 14th 2009, 04:28 PM
Jhevon
Quote:

Originally Posted by arbolis
lol, I can't believe I missed that. That's the change of coordinates. Believe it or not, I realized this a few minutes ago while cooking. (Nod)

Edit: I'm stuck on the last integral : $V=-\frac{4}{3} a^{\frac{3}{2}} \int _0^{\frac{\pi}{2}} \sin ^{\frac{3}{2}} (\theta) d\theta$.

how exactly did you end up with that integral? that guy never showed up when i did it
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