Describing a solid determined by a triple integral in spherical coordinates

ksdhart

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I'm having a bit of difficulty with a problem from my Calculus IV class. I already solved this problem, but it doesn't match the answer given, and there's no justification for why that's the answer. The problem states:

This iterated integral uses spherical coordinates. Describe the solid determined by the limits of integration.

0π20π2  01f(ρ,θ,ϕ)ρ2sin(ϕ)dρdθdϕ\displaystyle \displaystyle \int _0^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\:\:\int _0^1\:f\left(\rho ,\:\theta ,\:\phi \right)\rho ^2\sin \left(\phi \right)d\rho d\theta d\phi

My textbook says that, given a point in rectangular coordinates (x,y,z), Theta "measure the angular rotation from the x-axis to the ray containing the origin and the point (x,y,0)." and Phi "is the angle that the ray from the origin through point P makes with the z-axis."

So, rather than try and draw a 3D graph with perspective, I broke it apart into the xy-plane and yz-plane. In the xy-plane, the only coordinates that matter are Rho and Theta, and they can be I believe they can treated as the same as I would polar coordinates. Rho goes from 0 to 1 so we're dealing with a unit sphere, or some fraction thereof. Theta goes from 0 to pi/2. When Theta is 0, that's a straight line along the x-axis. And when Theta is pi/2, it's a straight line along the y-axis. Thus, I have a quarter circle in the first quadrant of the xy-plane. In the yz-plane, I applied similar reasoning. Based on the way my book draws their diagrams showing what angle Phi measures, I said that when Phi is 0, I'd have a straight line along the z-axis. And when Phi is pi/2, I'd have a straight line along the y-axis. Thus, I have a quarter circle in the first quadrant of the yz-plane.

And since I have a quarter circle in the first quadrants of both the xy-plane and the yz-plane, that suggests to me that the solid is a quarter of a sphere. However, the answer given is that it's a hemisphere. I don't have even the slightest clue how they might have come to that answer. Any help would be much appreciated.
 
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