Question

Where e is the solid hemisphere with the centre at the origin?

Answer 1

Okay, let’s do some integrals!


What? No… No they’re not! I’ll show you!

(Just like how you left all those drafts unfinished because you couldn’t figure out a simple integral in polar coordinates?)

But… That problem had trigonometry in it!

(Whatever.)

Pffft. Loser left brain is always skeptical.

Let’s define a coordinate system, (x, y, z) such that the hemisphere has its flat face on the xy plane.

Okay, we need to integrate over the region once, to find the z-coordinate of the center of mass. We already know the center of mass is over the origin by simple symmetry arguments.

Then we can uhh…

*Thinks a while*

*Looks at Pauls Online Math Notes*

(You’re embarrassing yourself again; I TOLD you to pay attention in math, but oh no it’s not important enough)

You’re making me nervous!

So, we can express the center of mass as:

mz=1M∭Vρ(x,y,z)zdVmz=1M∭Vρ(x,y,z)zdV

You can think of this formula as taking the average of the z-axis positions of each infinitesimally small particle, weighted according to the mass of the particle, which is represented by the density function, ρρ. But we’re assuming that the hemisphere is of uniform density, so we can take the constant function out of the integral:

mz=ρM∭VzdVmz=ρM∭VzdV. And we can then cancel out the density factor from the mass and plug in the volume of a hemisphere:

mz=32πR3∭VzdVmz=32πR3∭VzdV. Now we have a simple integral to evaluate in spherical coordinates:

mz=32πR3∫π20∫2π0∫R0r3cosϕsinϕdrdθdϕmz=32πR3∫0π2∫02π∫0Rr3cos⁡ϕsin⁡ϕdrdθdϕ, where rr represents the radius in spherical coordinates, ϕϕ is the angle between the point and the z-axis, and θθ is the azimuthal angle. I’ve already multiplied the Jacobian into the integrand.

(Because you Googled it at the last minute and realized you screwed up).

Zip it. Onwards, evaluating the innermost nested integral (and using a trigonometric identity to simplify),

mz=316πR3∫π20∫2π0R4sin(2ϕ)dθdϕmz=316πR3∫0π2∫02πR4sin⁡(2ϕ)dθdϕ.

Continuing outwards, we notice that the integrand is independent of θθ, so we multiply the outside by 2π2π,

mz=38R3∫π20R4sin(2ϕ)dϕmz=38R3∫0π2R4sin⁡(2ϕ)dϕ.

Evaluating the final integral,

mz=316R32R4mz=316R32R4.

(I saw WolframAlpha come up for that integral)

We’re almost finished!

mz=38Rmz=38R.

There’s our result! After all that math, we got that the center of mass of a hemisphere is 3/8 of the way up.

Answer 2

x2 + y2 + z2 dV , where E is the solid hemisphere that lies above the xy- plane and centers at the origin with radius 1