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Math Help - multiple integration: hemisphere/cylinder

  1. #1
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    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
    Last edited by headaches11; November 5th 2009 at 04:46 AM.
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  2. #2
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    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.
    Attached Thumbnails Attached Thumbnails multiple integration:  hemisphere/cylinder-cylinderproblem.jpg  
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  3. #3
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    Quote Originally Posted by shawsend View Post
    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?)
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  4. #4
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    Quote Originally Posted by headaches11 View Post

    \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)?
    Attached Thumbnails Attached Thumbnails multiple integration:  hemisphere/cylinder-circleplot.jpg  
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  5. #5
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    Quote Originally Posted by shawsend View Post
    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:



    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
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  6. #6
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    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
    Attached Thumbnails Attached Thumbnails multiple integration:  hemisphere/cylinder-halfcircle.jpg  
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