multiple integration: hemisphere/cylinder

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• Nov 5th 2009, 04:18 AM
headaches11
multiple integration: hemisphere/cylinder
Hi, I've been stuck on this problem for a while now. I would be grateful for any help please.?

Q. Find the surface area remaining when the hemisphere

$x^2 + y^2 +z^2 = a^2$

is cut by the cylinder $x^2 + y^2 = ax$

Thanks for reading
• Nov 5th 2009, 08:54 AM
shawsend
Isn't that the surface area remaining after I cut out the part over the red circle in the plot below?

Then that area is:

$2\pi a^2-\mathop\int\int\limits_{\hspace{-15pt}\text{Red}} \sqrt{f_x^2+f_y^2+1}dxdy$

where $f(x,y)=\sqrt{a^2-(x^2+y^2)}$

How about converting that to polar coordinates. I'm not absoutely sure, but I initially get:

$2\pi a^2-2\int_0^{\pi/2}\int_0^{a\cos(t)}\frac{a}{\sqrt{a^2-r^2}}rdrdt$

You try and see what integral you get when you make the conversion.
• Nov 5th 2009, 10:11 AM
headaches11
Quote:

Originally Posted by shawsend
Isn't that the surface area remaining after I cut out the part over the red circle in the plot below?

Then that area is:

$2\pi a^2-\mathop\int\int\limits_{\hspace{-15pt}\text{Red}} \sqrt{f_x^2+f_y^2+1}dxdy$

where $f(x,y)=\sqrt{a^2-(x^2+y^2)}$

How about converting that to polar coordinates. I'm not absoutely sure, but I initially get:

$2\pi a^2-2\int_0^{\pi/2}\int_0^{a\cos(t)}\frac{a}{\sqrt{a^2-r^2}}rdrdt$

You try and see what integral you get when you make the conversion.

Hi Shawsend, thanks for taking the time to look into my problem.

However, there are a few things I do not follow:

$\mathop\int\int\limits_{\hspace{-15pt}\text{Red}} \sqrt{f_x^2+f_y^2+1}dxdy$

I do not understand where the above expression has come from? (is it a well known formula?) and why have you used f(x,y) and not f(x,y,z) as we are in 3d.
does z= f(x,y)?

i.e $z^2 = a^2 - ( x^2 +y^2)$
therefore $z = sqrt(a^2 - ( x^2 +y^2))$... is this where your f(x,y) came from?

And what does the notation $f_x^2$ mean? (im assuming a derivative wrt x?)
• Nov 5th 2009, 10:41 AM
shawsend
Quote:

Originally Posted by headaches11

$\mathop\int\int\limits_{\hspace{-15pt}\text{Red}} \sqrt{f_x^2+f_y^2+1}dxdy$

I do not understand where the above expression has come from? (is it a well known formula?) and why have you used f(x,y) and not f(x,y,z) as we are in 3d.
does z= f(x,y)?

i.e $z^2 = a^2 - ( x^2 +y^2)$
therefore $z = sqrt(a^2 - ( x^2 +y^2))$... is this where your f(x,y) came from?

And what does the notation $f_x^2$ mean? (im assuming a derivative wrt x?)

Yeah, that's a standard formula for the area of a surface f(x,y), (which in this case is as you noted: $z=\sqrt{a^2-(x^2+y^2)}$) over a region R in the plane and that's partials with respect to x and y.

Also, in the conversion to polar coordinates, need to figure out r(t) (blue line in the plot below) as a function of it's angular value t as t goes from zero to $\pi/2$. You can do that right if the equation of that circle is $(x-a/2)^2+y^2=(a/2)^2$ and $x=r\cos(t)$ and $y=r\sin(t)$?
• Nov 5th 2009, 03:21 PM
headaches11
Quote:

Originally Posted by shawsend
Yeah, that's a standard formula for the area of a surface f(x,y), (which in this case is as you noted: $z=\sqrt{a^2-(x^2+y^2)}$) over a region R in the plane and that's partials with respect to x and y.

Also, in the conversion to polar coordinates, need to figure out r(t) (blue line in the plot below) as a function of it's angular value t as t goes from zero to $\pi/2$. You can do that right if the equation of that circle is $(x-a/2)^2+y^2=(a/2)^2$ and $x=r\cos(t)$ and $y=r\sin(t)$?

Hi shawsend, I have made a lot of progress since your last reply, but i am still having problems with the limits of integration.

You calculated:
http://www.mathhelpforum.com/math-he...74a4de71-1.gif

I have the same answer but I dont not know why you have 2*integral. (cannot find where your 2 came from?) I found the first limit for integrating over r ... $acos(t)$... by transforming the equation of the cylinder into polar coords, did some simplifying using trig identities.

The last problem is understanding how you find the limit when integrating over theta, or t in your case (0..pi/2) did not get the circle idea you gave me in the last post.

Cheers
• Nov 5th 2009, 04:39 PM
shawsend
I'm only integrating over half of the circle so I multiply by two since it's symmetric across the x-axis. That integration then gives the surface area of the sphere above the red circle. That half-area is shown below in light red. In order to integrate that via polar coordinates, I have to break up the half red circle in terms of r and theta. You'll have to review that in a calculus book to fully understand it. And if $(x-a/2)^2+y^2=(a/2)^2$ is the circle and $x=r\cos(t)$ and $y=r\sin(t)$, making that substitution, I get $r^2-a r \cos(t)=0$ or factoring out an r and disregarding the r=0 solution, I get $r=a \cos(t)$. That is the functional relationship I need in polar coordinates giving me the point on the circle as that line I drew in it goes from 0 to $\pi/2$. Then, since we only want the surface area on the upper hemisphere:

$S=2\pi a^2-2\mathop\int\int\limits_{\hspace{-20pt}\text{half-red}} \frac{a}{\sqrt{a^2-(x^2+y^2)}}dxdy$

Now we make the change to polar coordinates noting that $dxdy=rdrdt$ and $r^2=x^2+y^2$ and we've just obtained the limits so we get:

$S=2\pi a^2-2\mathop\int\int\limits_{\hspace{-20pt}\text{half-red}} \frac{a}{\sqrt{a^2-(x^2+y^2}}dxdy=2\pi a^2-2\int_0^{\pi/2}\int_0^{a\cos(t)} \frac{a}{\sqrt{a^2-r^2}}rdrdt$