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Math Help - Stuck on this problem

  1. #1
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    Stuck on this problem

    I've included a picture stating the problem. How do I approach this?



    I recognize that x^2 + y^2 is the equation of a circle, but that's about all I have.
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  2. #2
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    Quote Originally Posted by Ares_D1 View Post
    I've included a picture stating the problem. How do I approach this?



    I recognize that x^2 + y^2 is the equation of a circle, but that's about all I have.
    When you rotate the curve z = e^{-x^2} about the z axis you get the surface z = e^{-r^2} (in cylindrical polar coords). Since x = r \cos \theta, y = r \sin \theta then r^2 = x^2 + y^2, giving your answer.
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  3. #3
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    Quote Originally Posted by Danny View Post
    When you rotate the curve z = e^{-x^2} about the z axis you get the surface z = e^{-r^2} (in cylindrical polar coords). Since x = r \cos \theta, y = r \sin \theta then r^2 = x^2 + y^2, giving your answer.
    Thanks for the help Danny, but I'm not sure I follow that last bit. Could you explain further how x = r \cos \theta and y = r \sin \theta , and how, from that, you end up with r^2 = x^2 + y^2?
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  4. #4
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    Quote Originally Posted by Ares_D1 View Post
    Thanks for the help Danny, but I'm not sure I follow that last bit. Could you explain further how x = r \cos \theta and y = r \sin \theta , and how, from that, you end up with r^2 = x^2 + y^2?
    x = r \cos \theta and y = r \sin \theta are polar coordinates - an alternate way to describe a point in 2D. Since

     <br />
\sin ^2 \theta + \cos ^2 \theta = 1, \text{then }\;x^2 +y^2 = r^2\left(\cos ^2 \theta + \sin ^2 \theta \right) = r^2<br />
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  5. #5
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    By the way, the statement " z= e^{x^2} lies in the xy-plane" is incorrect. Since you are then to rotate around the z-axis, I imagine it was supposed to be "lies in the xz-plane".
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