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Surface integrals

  1. Jan 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the area of the part of z^2=xy that lies inside the hemisphere x^2+y^2+z^2=1, z>0

    2. Relevant equations
    da= double integral sqrt(1+(dz/dx)^2+(dz/dy)^2))dxdy

    3. The attempt at a solution
    (dz/dx)^2=y/2x
    (dz/dy)^2=x/2y
    => double integral (x+y)(sqrt(2xy)^-1/5) dxdy

    Now I'm guessing that a change of coordinates will be useful here. I was thinking spherical coordinates, due to the presence of a sphere. But I'm not too sure?

    many thanks :)
     
  2. jcsd
  3. Jan 23, 2016 #2

    Ray Vickson

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    You have made a good start. What is stopping you from going ahead? It is much too early to guess about what kind of coordinate transformation to make: write down the complete details of your final double integral first, then decide on any possible changes of variables.
     
  4. Jan 23, 2016 #3
    I thought I had? I don't see how I can take that integral any further?
    Thank you so much for replying- I really appreciate it :)
     
  5. Jan 23, 2016 #4

    Ray Vickson

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    But: you are only half-finished. One of the most important aspects (that you have not yet addressed) is: what is the integration region in the (x,y)-plane? Without that, you can go nowhere.

    Besides: your expression for dA is incorrect: it is not ##(x+y)/\sqrt{2xy} \: dx \, dy##.
     
  6. Jan 23, 2016 #5
    Ohh of course. So would the limits be x=-sqrt(1-y^2) to x=sqrt(1-y^2) and y from -1 to 1?

    Is it not?

    I got dz/dx=y/sqrt(2xy) and dz/dy=x/sqrt(2xy)

    so 1+(dz/dx)^2+(dz/dy)^2= 1+y/2x+x/2y = (2xy+y^2+x^2)/2xy = (x+y)^2/2xy

    so sqrt(1+(dz/dx)^2+(dz/dy)^2) = (x+y)/sqrt(2xy) ?

    I am known for making a lot of algebraic mistakes so its more than possible that I've don't this wrong, but I can't see where

    Thank you :)
     
  7. Jan 23, 2016 #6

    Ray Vickson

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    Your region is still incorrect. Instead of writing down the first thing that pops into your mind, take the time to it out carefully and in detail from first principles. (1) What determines the integration region in (x,y) - space? (2) How would you express this algebraically? (3) You will get a boundary curve in (x,y)-space. In terms of its algebraic characterization, what type of curve will you get?

    You may need to devote quite a bit of time and effort to this part of the problem, in which case there are no shortcuts. It is more important to get it right than to save some time.

    I would bet that you have worked out such problems before, in other courses, or maybe earlier in the current course.

    And no: you still do not have the correct expression for dA. One last hint: your derivatives ##\partial z/\partial x## and ##\partial z/\partial y## are wrong.
     
  8. Jan 24, 2016 #7
    I'm really struggling to think of how to do that though.

    And again, I'm really sorry but I don't see how? z=sqrt(2xy) dz/dx=1/2*(2y)*(2xy)^-1/2 = y/(2xy)^1/2 ? or have I made another mistake that I just can't see
     
  9. Jan 24, 2016 #8

    vela

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    You wrote ##z^2 = xy## in the original post. Now you're saying ##z = \sqrt{2xy}##. Which is correct?
     
  10. Jan 24, 2016 #9
    Ahh my mistake. It was meant to be z^2=2xy in the original post. Sorry!!
     
  11. Jan 24, 2016 #10

    Ray Vickson

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    That changes everything! It changes the (x,y)-region, it changes the integrand---everything.
     
  12. Jan 24, 2016 #11
    I'm really sorry :( I didn't mean to mistype it- i'm just not that good with computers.

    But I'm lost as to where to go from here....
     
  13. Jan 24, 2016 #12

    Ray Vickson

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    All my previous advice still holds: (1) work out the (x,y)-region carefully, from first principles. (2) Double-check your x and y-derivatives of ##z = \sqrt{2} \sqrt{xy}##. (3) Write down in detail the integration involved in getting your final answer. (4) Then, and only then, worry about how to actually do the integral or integrals.

    Beyond that, I am not permitted to say more, at least by my interpretation of the PF rules.
     
  14. Jan 26, 2016 #13
    Its fine I managed to work it out. But thank you for your help anyway :)
     
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