Question

Multivariable integration help:

Compute the moment of inertia around the y-axis associated with the region |y| 12 years ago

Answer 1

the moment of inertia \((I_x,I_y)\) is computed using:$$I_x=\iint_R x^2 \delta(x,y)\ dx\ dy\\I_y=\iint_R y^2\delta(x,y)\ dx\ dy$$

Answer 2

here \(R\) is given by \(-x^2<y<x^2\) and \(-2<x<2\) whereas \(\delta=5/32\) ergo:$$\begin{align*}I_x&=\int_{-2}^2\int_{-x^2}^{x^2}x^2\cdot\frac5{32}\ dy\ dx\\&=\frac5{32}\int_{-2}^2 x^2\int_{-x^2}^{x^2}\ dy\ dx\\&=\frac5{32}\int_{-2}^2 2x^4\ dx\\&=\frac18\int_0^25x^4\ dx\\&=\frac18(2^5-0^5)\\&=4\end{align*}$$

Answer 3

can you find \(I_y\)?

Answer 4

oh duh...I forgot I had 2 dimensions to work with. Thanks a ton!

Answer 5

no problem :-p it's the same integrand only \(y^2\)

Answer 6

wait so same as the first line in the bounds too?

Answer 7

indeed

Answer 8

just a sec, somewhat confused, are the bounds changed to y^2? no, right?

Answer 9

nope!

Answer 10

ok got the answers. you were great. Saved my neck here.

Answer 11

$$I_y=\int_{-2}^2\int_{-x^2}^{x^2}y^2\cdot\frac5{32}\ dy\ dx=\frac5{32}\int_{-2}^2\left[\frac13y^3\right]_{-x^2}^{x^2}\ dx=\frac5{32}\int_{-2}^2\frac23x^6dx$$so we get:$$I_y=\frac5{48}\int_{-2}^2 x^6\ dx=\frac5{24}\int_0^2x^6\ dx=\frac5{24}\cdot\frac17(2^7-0^7)=\frac{5\cdot16}{3\cdot7}=\frac{80}{21}$$