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I'm trying to integrate this volume in spherical and cylindrical coordinates, but having difficulty finding my bounds of integration;

I'm given the region in the first octant bounded by z = $\sqrt{x^2+y^2}$, z = $\sqrt{1-x^2-y^2}$, y = x and y = $\sqrt{3}x$ and I need to evaluate $\iiint_V{}dV$

When proceeding to integrate with spherical and cylindrical coordinates I am not getting the right bounds such that both methods equate to the same volume? I am definitely missing something. Any and all advice would be much appreciated!

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This region seems better defined using spherical coordinates than cylindrical. We are given that the region is between two vertical planes $y=x$ and $y=\sqrt3x$, and it is between the sphere $x^2 + y^2 + z^2 = 1$ and the upper half of the cone $x^2 + y^2 = z^2$. From this, we can set the bounds to be: $$ \frac\pi3 \le \theta \le \frac\pi4$$ from the region of angles between the two lines (arctan of root(3) is pi/3) $$0 \le \phi \le \frac\pi4 $$from the intersection of the cone and sphere $$0 \le r \le 1 $$from the radius of sphere

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  • $\begingroup$ Hope this helps! If you need any clarification just ask :D $\endgroup$ – Akarsh Verma Nov 28 '16 at 2:46
  • $\begingroup$ Thank you very much! However can you help me by explaining your reasoning for the bounds of $\phi$ $\endgroup$ – Nelly Nov 28 '16 at 3:02
  • $\begingroup$ There's something crucial I forgot in the statement of the problem which I just added; I forgot to mention that I'm integrating in the first octant of the specified region. That changes the bounds of integration right? $\endgroup$ – Nelly Nov 28 '16 at 3:47
  • $\begingroup$ I had actually already assumed you were in the first octant by accident. Whoops! But have you learned spherical coordinates? If so, $\phi$ is the angle from the north pole in the clockwise direction that you are integrating. In order to describe the region between the cone and sphere, I am going from the north pole and then until I get the cone (intersection of cone and sphere) $\endgroup$ – Akarsh Verma Nov 28 '16 at 4:00
  • $\begingroup$ If the question was not only first octant, I would also have to take into account the other region of theta. The region between the lines y = sqrt(3) and y = x in quadrant 3 in the xy coordinate plane is between two different angles. Theta being set as I did above only accounts for the first quadrant $\endgroup$ – Akarsh Verma Nov 28 '16 at 4:21

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