anonymous
  • anonymous
Can someone help me solve this iterated integral? Tanks.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Here's the equation.
1 Attachment
anonymous
  • anonymous
the one marked red?
anonymous
  • anonymous
Yes, xD

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anonymous
  • anonymous
use polar
anonymous
  • anonymous
substitutions
anonymous
  • anonymous
I have a test in 4days time on this stuff, double/triple integrals etc
anonymous
  • anonymous
Oh, okay, thanks. Good luck to you! I'll get back here if I can't do it lol.
anonymous
  • anonymous
let A be the angle
anonymous
  • anonymous
x=rcos(A) y= rsin(A)
anonymous
  • anonymous
the jacobian of the substitution is "f"
anonymous
  • anonymous
edit "r"
anonymous
  • anonymous
how would I get r? is it = 1? I figured out theta, which is from 0 to pi/4
anonymous
  • anonymous
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anonymous
  • anonymous
not the best picture lol
anonymous
  • anonymous
lol, i'll upload mine
anonymous
  • anonymous
the angle goes from 90degrees to 45degrees actually
anonymous
  • anonymous
but the difference is still the same
anonymous
  • anonymous
and our r goes from 0<=r<=1
anonymous
  • anonymous
because 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
  • anonymous
This is what I came up with, is it correct?
1 Attachment
anonymous
  • anonymous
Elec, your drawing is correct, but the sector described is actually the portion you have marked 45 degrees. Good job
anonymous
  • anonymous
I got it based on the 2 iterated integrals below in pic. ok, thx, i think i can do it now, 0<=r<=1
anonymous
  • anonymous
wait, the region I drew was correct I am fairly sure
anonymous
  • anonymous
I know it doesnt impact the final answer , but the integral has x as the lower limit, and the semicircle as the upper limit
anonymous
  • anonymous
is it 2y+2x
anonymous
  • anonymous
\[= \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta \int\limits_{0}^{1} r^2 dr \]
anonymous
  • anonymous
= \[\frac{\pi}{4} \times \frac{1}{3} = \frac{\pi}{12}\]
anonymous
  • anonymous
What about the region I drew? Is that acceptable?
anonymous
  • anonymous
Oh yeah, you're right, I was looking at it cross ways. Choose some x, and the y goes from x to the semicircle.
anonymous
  • anonymous
Revan, 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
  • anonymous
But isn't it, 0<=y<=x?
anonymous
  • anonymous
no!
anonymous
  • anonymous
the region you drew was y<=x and y<=sqrt(1-x^2)
anonymous
  • anonymous
the intersection of those region , thats what your picture shows
anonymous
  • anonymous
even 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
  • anonymous
Out of curiosity, revan, where did you get sq rt 2/2 from?
anonymous
  • anonymous
http://openstudy.com/updates/attachments/4dcf65a1dfd28b0b901b0dfd-revangale-1305438547251-photo1.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
  • anonymous
Part of confusion might be: that's not the question you posted here for this post.
anonymous
  • anonymous
Lol, sorry >.< The question I posted is the single iterated integral. I just needed help solving it.
anonymous
  • anonymous
yes, the way you did was very wrong lols
anonymous
  • anonymous
:D thats why i posted lol
anonymous
  • anonymous
\[\int\limits(ax+b)^n = \frac{(ax+b)^{n+1} } {a(n+1)} +C \]
anonymous
  • anonymous
only works for linear functions
anonymous
  • anonymous
raised to a power
anonymous
  • anonymous
Now elec, is writing wrong thing
anonymous
  • anonymous
You said to do poloar coordinates. So, I think I can solve it now. I'll post it when I'm done.
anonymous
  • anonymous
Btw, y is r from 0 to 1? o.0
anonymous
  • anonymous
nvm . ..
anonymous
  • anonymous
Revan, if given this from scratch, you choose polar, but if given the integrals already compiled it is very easy to read.
anonymous
  • anonymous
Yea, but I cant' solve it w/o using polar, but either system, the answer will be the same?
anonymous
  • anonymous
What do you mean by 'solve'
anonymous
  • anonymous
or find volume? that's what you mean?
anonymous
  • anonymous
http://openstudy.com/updates/attachments/4dcf65a1dfd28b0b901b0dfd-revangale-1305438547251-photo1.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
  • anonymous
So, 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
  • anonymous
Well, 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
  • anonymous
Lol, okay, thx., yea I didn't know wat was going on.
anonymous
  • anonymous
BTW 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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