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

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oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1the moment of inertia \((I_x,I_y)\) is computed using:$$I_x=\iint_R x^2 \delta(x,y)\ dx\ dy\\I_y=\iint_R y^2\delta(x,y)\ dx\ dy$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1here \(R\) is given by \(x^2<y<x^2\) and \(2<x<2\) whereas \(\delta=5/32\) ergo:$$\begin{align*}I_x&=\int_{2}^2\int_{x^2}^{x^2}x^2\cdot\frac5{32}\ dy\ dx\\&=\frac5{32}\int_{2}^2 x^2\int_{x^2}^{x^2}\ dy\ dx\\&=\frac5{32}\int_{2}^2 2x^4\ dx\\&=\frac18\int_0^25x^4\ dx\\&=\frac18(2^50^5)\\&=4\end{align*}$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1can you find \(I_y\)?

alffer1
 one year ago
Best ResponseYou've already chosen the best response.0oh duh...I forgot I had 2 dimensions to work with. Thanks a ton!

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1no problem :p it's the same integrand only \(y^2\)

alffer1
 one year ago
Best ResponseYou've already chosen the best response.0wait so same as the first line in the bounds too?

alffer1
 one year ago
Best ResponseYou've already chosen the best response.0just a sec, somewhat confused, are the bounds changed to y^2? no, right?

alffer1
 one year ago
Best ResponseYou've already chosen the best response.0ok got the answers. you were great. Saved my neck here.

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1$$I_y=\int_{2}^2\int_{x^2}^{x^2}y^2\cdot\frac5{32}\ dy\ dx=\frac5{32}\int_{2}^2\left[\frac13y^3\right]_{x^2}^{x^2}\ dx=\frac5{32}\int_{2}^2\frac23x^6dx$$so we get:$$I_y=\frac5{48}\int_{2}^2 x^6\ dx=\frac5{24}\int_0^2x^6\ dx=\frac5{24}\cdot\frac17(2^70^7)=\frac{5\cdot16}{3\cdot7}=\frac{80}{21}$$
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