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

is cut by the cylinder

Thanks for reading

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- November 5th 2009, 03:18 AMheadaches11multiple 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

is cut by the cylinder

Thanks for reading - November 5th 2009, 07:54 AMshawsend
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:

where

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

You try and see what integral you get when you make the conversion. - November 5th 2009, 09:11 AMheadaches11

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

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

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

therefore ... is this where your f(x,y) came from?

And what does the notation mean? (im assuming a derivative wrt x?) - November 5th 2009, 09:41 AMshawsend
Yeah, that's a standard formula for the area of a surface f(x,y), (which in this case is as you noted: ) 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 . You can do that right if the equation of that circle is and and ? - November 5th 2009, 02:21 PMheadaches11

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 ... ... 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 - November 5th 2009, 03:39 PMshawsend
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 is the circle and and , making that substitution, I get or factoring out an r and disregarding the r=0 solution, I get . 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 . Then, since we only want the surface area on the upper hemisphere:

Now we make the change to polar coordinates noting that and and we've just obtained the limits so we get: