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
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
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.
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?)
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 ?
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:
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
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: