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anonymous
 5 years ago
Can someone help me solve this iterated integral? Tanks.
anonymous
 5 years ago
Can someone help me solve this iterated integral? Tanks.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have a test in 4days time on this stuff, double/triple integrals etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, okay, thanks. Good luck to you! I'll get back here if I can't do it lol.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the jacobian of the substitution is "f"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how would I get r? is it = 1? I figured out theta, which is from 0 to pi/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not the best picture lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, i'll upload mine

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the angle goes from 90degrees to 45degrees actually

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the difference is still the same

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and our r goes from 0<=r<=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because we are inside the disc x^2 +y^2 <1 , so [rcos(A)]^2 +[ rsin(A)]^2 <1 , r^2<1 but r is also positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is what I came up with, is it correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Elec, your drawing is correct, but the sector described is actually the portion you have marked 45 degrees. Good job

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got it based on the 2 iterated integrals below in pic. ok, thx, i think i can do it now, 0<=r<=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, the region I drew was correct I am fairly sure

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know it doesnt impact the final answer , but the integral has x as the lower limit, and the semicircle as the upper limit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[= \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta \int\limits_{0}^{1} r^2 dr \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0= \[\frac{\pi}{4} \times \frac{1}{3} = \frac{\pi}{12}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What about the region I drew? Is that acceptable?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh yeah, you're right, I was looking at it cross ways. Choose some x, and the y goes from x to the semicircle.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Revan, elec's drawing is right. Also you have a sq rt 2/2 in there, that I think you got from your other problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But isn't it, 0<=y<=x?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the region you drew was y<=x and y<=sqrt(1x^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the intersection of those region , thats what your picture shows

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0even though they would both give the same answer it is very likely you would lost a mark or two , because your drawing and the limits of the angle dont reflect the correct region

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Out of curiosity, revan, where did you get sq rt 2/2 from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/updates/attachments/4dcf65a1dfd28b0b901b0dfdrevangale1305438547251photo1.jpg from the two integrals at the bot of the pic. I have to sketch Region of the sum of the two and then replace the sum by an equivalent single iterated integral and evalute

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Part of confusion might be: that's not the question you posted here for this post.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol, sorry >.< The question I posted is the single iterated integral. I just needed help solving it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, the way you did was very wrong lols

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:D thats why i posted lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits(ax+b)^n = \frac{(ax+b)^{n+1} } {a(n+1)} +C \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0only works for linear functions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now elec, is writing wrong thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You said to do poloar coordinates. So, I think I can solve it now. I'll post it when I'm done.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Btw, y is r from 0 to 1? o.0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Revan, if given this from scratch, you choose polar, but if given the integrals already compiled it is very easy to read.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yea, but I cant' solve it w/o using polar, but either system, the answer will be the same?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you mean by 'solve'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or find volume? that's what you mean?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/updates/attachments/4dcf65a1dfd28b0b901b0dfdrevangale1305438547251photo1.jpg I'm supposed to combine the sum of the two iterated integrals at the bottom of this pic into a single iterated integral and evalute it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, I drew the graph of the two iterated integrals together, to get the overall Region, and then use the overall Region to set up a single iterated integral. But, I couldn't solve it, unless it was in polar system. Is this the right approach?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, just looking at the picture, it is a sector of a circle and it's a polar type thing. So sure. It is just a kind of trick question, your teacher wants to know if you understand what's going on

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol, okay, thx., yea I didn't know wat was going on.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0BTW ele engineer is right in using transformation but going from rectangular coordinates to polar is something that should be easy we know da> r dr dtheta where x= r cos theta y = r sin theta and x^2 + y^2 = r ^2 and if you actually draw it out its easier to know the boundaries.
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