This is easy enough to draw, it is just a wedge and 1/8 of a sphere centered at the origin. The mass will be given by: $$m= \int \int \int p(x,y,z)dV$$ I just need to figure out the bounds of the sphere, but that is where I get confused.
- Which coordinate system could/will work? Could I use rect, cyl, or spherical?
- Could I first figure it out in rectangular and then transform it any of the other systems? I haven't learned transformations yet.
If I use spherical, will my bounds be as follows? $$0 \le p \le5 \;\;\;\; 0 \le \theta \le {\pi\over2} \;\;\;\; 0 \le\phi \le{\pi \over 2}$$ and thus: $$\int_0^{\pi \over 2} \int_0^{\pi \over 2} \int_0^5{(z)}p^2\sin\phi dp d \theta d \phi$$?
Thanks for any help, I have a hard time getting these problems figured out.
EDIT: I just noticed that z will need to be converted, but I'm still concerned about the bounds.
$\endgroup$1 Answer
$\begingroup$You are using spherical coordinates $$\int_0^{\pi \over 2} \int_0^{\pi \over 2} \int_0^5{(z)}p^2\sin\phi dp d \theta d \phi$$
Thus you need to substitute $pcos (\phi)$ for z in your integrand.
The new integral is $$\int_0^{\pi \over 2} \int_0^{\pi \over 2} \int_0^5 p^3 \cos(\phi)\sin\phi dp d \theta d \phi$$
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