Question

@ganeshie8

Answer 1

I always like to tag ganeshie

Answer 2

lol ur tags always dread me.. :/

Answer 3

@ganeshie8 is too modest.

Answer 4

Ok, so Let H be a solid hemisphere of radius a whose density at any point is proportional to its distance from the center of the base.
Find the mass H, there's more but just need the set up really

Answer 5

So density = kp

Answer 6

Eugh densities... doyou have to find the center of mass and stuff...

Answer 7

yeah but w e lets just find mass

Answer 8

Also asks for moment of inertia ahaha

Answer 9

Well, theres \(M_x\) and \(M_y\)

Answer 10

All I remember is that \(M_x\) has to do with y, and \(M_y\) has to do with x.

Answer 11

mass = density * volume

Answer 12

agree

Answer 13

You just apply the formula for both of those.

Answer 14

just setup a triple integral over H for the density

Answer 15

p is from 0 to a then?

Answer 16

Oh wait, from what ganeshie is saying..

Answer 17

theta 0 to 2 pi

Answer 18

phi though

Answer 19

\[M= \rho A = \rho(x,y)\cdot \Delta x \cdot \Delta y\]

Answer 20

should be easy after doing that previous hard problem..

Answer 21

Right?

Answer 22

I remembered this because area across a solid volume is always with respect to both change in x and y.

Answer 23

\[M = \iint_R\rho(x,y)dxdy\]

Answer 24

that works for a solid without thickness..

we need to add another dimension for thickness...

Answer 25

Oh... haha, that'sz what I was missing.

Answer 26

The `thickness` as in constant value \(k\).

Answer 27

\[M = \iint_Rk\rho(x,y)dxdy\]

Answer 28

\[M = \iint_R\rho(x,y,z)dxdydz\]

here \(\rho(x,y,z)\) is the density function.. not to be confused with distance from origin in spherical coordinates..

Answer 29

** \[M = \iiint_R\rho(x,y,z)dxdydz\]

Answer 30

\[m=\int\limits_{0}^{2\pi} \int\limits_{0}^{\pi/2} \int\limits_{0}^{a} kp ...etc\]

Answer 31

So close, damn.

Answer 32

This is a chump question, I got a better

Answer 33

\[m=\int\limits_{0}^{2\pi} \int\limits_{0}^{\pi/2} \int\limits_{0}^{a} (k\rho)~ \rho^2\sin\phi d\rho d\phi d\theta \]

Answer 34

ganeshie get ready to be tagged