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Math Help - triple integral

  1. #1
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    Exclamation triple integral

    evaluate the triple integral by switching to cylindrical coordinates.
    the triple integral of 1+ (x)/(sqrt(x^2+y^2))
    over S which is the solid bounded by the paraboloids z=x^2+y^2 and z=1-x^2-y^2

    i changed the equation to cylindrical coordinates and simplified to get 1+cos(theta)
    my main problem is finding the boundaries to integrate from.
    does anyone know how i can find the boundaries for r, theta, and z?

    the final answer in the back of the book is pi/4 if that helps.
    thank you so much! i just can't figure out how to find the boundaries when it's not in rectangular coordinates..
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  2. #2
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    First, we find out where the right-side-up paraboloid z=x^2+y^2 and the upside-down paraboloid z=1-x^2-y^2 intersect. In doing this, we switch to cylindrical coordinates:

    \begin{aligned}<br />
x^2+y^2&=1-x^2-y^2\\<br />
r^2&=1-r^2\\<br />
2r^2&=1\\<br />
r^2&=\frac{1}{2}\\<br />
r&=\frac{1}{\sqrt{2}}.<br />
\end{aligned}

    The paraboloids therefore intersect in a circle of radius \frac{1}{\sqrt{2}}. As the region extends from z=0 to z=1, we may set up our integral as follows, remembering to include the extra factor r:

    \int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2}}}\int_0^1 (1+\cos \theta)r\,dz\,dr\,d\theta.

    Incidentally, I did not obtain \frac{\pi}{4} when I worked out this problem.
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