## alffer1 one year ago Multivariable integration help: Compute the moment of inertia around the y-axis associated with the region |y|<x^2, |x|<2, assuming constant density delta = 5/32. Not sure what to start with...

1. oldrin.bataku

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

2. oldrin.bataku

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*}

3. oldrin.bataku

can you find $$I_y$$?

4. alffer1

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

5. oldrin.bataku

no problem :-p it's the same integrand only $$y^2$$

6. alffer1

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

7. oldrin.bataku

indeed

8. alffer1

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

9. oldrin.bataku

nope!

10. alffer1

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

11. oldrin.bataku

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