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Math Help - Double integral surface area with polar coordinates

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    Double integral surface area with polar coordinates

    Problem: Find the surface area of the paraboloid z=4-x^2-y^2 that lies above the xy-plane. Finding where the function crosses the xy plane- 4=x^2+y^2 taking the gradient- \{2x,2y\} Useing the surface area formula- <br> <br>\int\int\sqrt{(4x^2+4y^2+1)}dxdy<br><br>This is where I'm unsure of how to continue. It seems that I could just set my bounds to be [-2,2] and [0,2] but this gives and incorrect answer.

    The solution book says to convert the function to polar coordinates as such -
    <br><br>
    \int_{0}^{2\pi}\int_{0}^{2}\sqrt{(4r^2+1)}drd\thet  a,<br><br>I'm unsure of the bounds for theta, wouldn't 0-2\pi give the surface area of the whole sphere and not the area above xy? How would you compute the total surface area if this equation just gives it to you above the xy-plane?<br>
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  2. #2
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    Quote Originally Posted by Reskal View Post
    Problem: Find the surface area of the paraboloid z=4-x^2-y^2 that lies above the xy-plane. Finding where the function crosses the xy plane- 4=x^2+y^2 taking the gradient- \{2x,2y\} Useing the surface area formula-

    \int\int\sqrt{(4x^2+4y^2+1)}dxdy

    This is where I'm unsure of how to continue. It seems that I could just set my bounds to be [-2,2] and [0,2] but this gives and incorrect answer. [That is because you would then be calculating the surface area over the square \color{red}-2\leqslant x\leqslant2,\ -2\leqslant y\leqslant2.]

    The solution book says to convert the function to polar coordinates as such -

    \int_{0}^{2\pi}\int_{0}^{2}\sqrt{(4r^2+1)}drd\thet  a, [Correct except that \color{red}drd\theta should be \color{red}r\,drd\theta.]

    I'm unsure of the bounds for theta, wouldn't 0-2\pi give the surface area of the whole sphere [It's not a sphere, it's a paraboloid, and you're calculating the area of that part of its surface that lies above the disk \color{red}x^2+y^2\leqslant4, which in polar coordinates corresponds to \color{red}0\leqslant\theta\leqslant2\pi,\ 0\leqslant r\leqslant2.] and not the area above xy? How would you compute the total surface area if this equation just gives it to you above the xy-plane?
    ..
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