Question

Find the centroid. Let W be the region inside the surfaces z=x^2+y^2 and x^2+y^2+z^2 = 2

Please help! I had enough trouble with centroids when they were in 2 dimensions, much less three, haha.

Answer 1

First we must notice that the two surfaces meet at z = 1. We must parametrize the solid in two parts: for \[0 \leq z \leq 1\] it's a paraboloid and for \[1 \leq z \leq \sqrt{2}\] it's a sphere. I used cylindrical coordinates and got the following result:
\[D = D_1 \cup D_2\] with \[D_1 = \{(r \cos{\theta}, r \sin{\theta}, z) \colon \theta \in [0, 2\pi), z \in [0, 1], r \in [0, \sqrt{z}]\}\] and\[D_2 = \{(r \cos{\theta}, r \sin{\theta}, z) \colon \theta \in [0, 2\pi), z \in [1, \sqrt{2}], r \in [0, \sqrt{2 - z^2}]\}.\]
It's easy to see that the centroid must lie somewhere in the Z axis (because of the obvious symmetries in the solid), so it's a point of the form \[(0, 0, C_z).\]

To find \[C_z\] we must use the definition: \[C_z = \frac{1}{M} \int_D z dm\] where \[dm\] is the mass element \[\rho dV.\] We can safely assume constant density \[\rho = 1\] (or any other number), so let's find the mass of the object:
\[M = \int_D dV = \int_0^{2\pi} \int_0^1 \int_0^\sqrt{z} dV + \int_0^{2\pi} \int_1^\sqrt{2} \int_0^\sqrt{2 - z^2} dV \]
The volume element in cylindrical coordinates is \[dV = r drdzd\theta\] so we evaluate those integrals and get \[M = \frac{1}{6}(8\sqrt{2} - 7)\pi.\]

Finally, we evaluate the previous integral to find

\[C_z = \frac{1}{M} \left( \int_0^{2\pi} \int_0^1 \int_0^\sqrt{z} z dV + \int_0^{2\pi} \int_1^\sqrt{2} \int_0^\sqrt{2 - z^2} z dV \right) = \frac{7}{2(8\sqrt{2} - 7)}\]
so the centroid is \[C = \left(0, 0, \frac{7}{2(8\sqrt{2} - 7)}\right).\]

Answer 2

genius!