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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...
 3 months ago
 3 months 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...
 3 months ago
 3 months ago

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

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

oldrin.batakuBest ResponseYou've already chosen the best response.1
can you find \(I_y\)?
 3 months ago

alffer1Best ResponseYou've already chosen the best response.0
oh duh...I forgot I had 2 dimensions to work with. Thanks a ton!
 3 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
no problem :p it's the same integrand only \(y^2\)
 3 months ago

alffer1Best ResponseYou've already chosen the best response.0
wait so same as the first line in the bounds too?
 3 months ago

alffer1Best ResponseYou've already chosen the best response.0
just a sec, somewhat confused, are the bounds changed to y^2? no, right?
 3 months ago

alffer1Best ResponseYou've already chosen the best response.0
ok got the answers. you were great. Saved my neck here.
 3 months ago

oldrin.batakuBest 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:$$I_y=\frac5{48}\int_{2}^2 x^6\ dx=\frac5{24}\int_0^2x^6\ dx=\frac5{24}\cdot\frac17(2^70^7)=\frac{5\cdot16}{3\cdot7}=\frac{80}{21}$$
 3 months ago
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