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Write the triple integral of the right circular cone of height $h$ using rectangular, cylindrical, and spherical coordinates. (The part of the cone $z=\sqrt{x^2+y^2}$ lying below the plane $z=h$ above the $xy$-plane).

Are these the correct?

Rectangular:

$\{ (x,y,z): \sqrt{x^2+y^2}\leq z \leq h, -\sqrt{h-y^2}\leq x \leq \sqrt{h-y^2}, -h\leq y\leq h \}$

Cylindrical:

$\{(r,\theta, z): 0 \leq r \leq h, 0 \leq \theta \leq 2\pi , r \leq z \leq h\} $

Spherical:

$\{ (\rho , \theta , \phi ): 0 \leq \rho \leq \sqrt{2}h, 0\leq \theta \leq 2\pi , 0 \leq \phi \leq \frac{\pi}{4} \} $

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    $\begingroup$ The formulas for the bounds depend on the order in which you integrate over the three variables. The bounds for rectangular coordinates look correct, but there is another order of integration that I think is simpler. The bounds for spherical coordinates cannot all be constants; what you wrote makes the "base" of the cone a spherical cap rather than a flat disk. $\endgroup$ – David K Apr 20 '16 at 22:31
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Rectangular:

$(x,y,z): \sqrt{x^2+y^2}<z<h, -h<y<h, -\sqrt{h-y^2}<y<\sqrt{h-y^2}$

Spherical:

$(\rho,\theta,\phi): 0<\rho<sec(\phi)h, 0<\theta<2\pi, 0<\phi<\frac{\pi}/{4}$

And you are correct with cylindrical coordinates

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  • $\begingroup$ I suggest to check the rectangular coordinates. $\endgroup$ – David K Apr 20 '16 at 22:31

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