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tdabboud Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{\pi/2} \int\limits_{0}^{\cos \theta} r^3 drd \theta\]
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.0
So integrate r first. What is \[\int_0^{cos \theta} r^3 dr\]
 3 years ago

tdabboud Group TitleBest ResponseYou've already chosen the best response.0
Thats the thing I am getting stuck right after that so I have this: \[\int\limits_{0}^{\pi/2}[[\cos \theta]^4/4  1/4] d \theta\]
 3 years ago

tdabboud Group TitleBest ResponseYou've already chosen the best response.0
taking the integral of that before evaluating seems to be some long number however
 3 years ago

chaguanas Group TitleBest ResponseYou've already chosen the best response.0
I think you need a little more work. cos theta shouldn't be an end point. As part of the set up you should evaluate cos theta and come up with some number for example cos pi/4 is square root 2
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.0
It should be just \[\int_0^{\pi/2}[\frac{1}{4}cos^4\theta ]d\theta\]
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.0
Evaluating \[\frac{r^4}{4}\] at r=0 does not give you 1/4
 3 years ago

tdabboud Group TitleBest ResponseYou've already chosen the best response.0
wow I am dumb, I was evaluating the cos again, okay you were right with the last statement
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.0
And for integrating cos^4 I think you need to make use of the half angle formula.
 3 years ago

tdabboud Group TitleBest ResponseYou've already chosen the best response.0
or could you just uses u substitution?
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.0
You can try, but I don't think it'll work out as nicely.
 3 years ago

tdabboud Group TitleBest ResponseYou've already chosen the best response.0
okay well thanks for your help atleast I got passed that one error I should be able to get it from here
 3 years ago
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