Can someone help me solve this iterated integral? Tanks.

- anonymous

Can someone help me solve this iterated integral? Tanks.

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- anonymous

Here's the equation.

##### 1 Attachment

- anonymous

the one marked red?

- anonymous

Yes, xD

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## More answers

- anonymous

use polar

- anonymous

substitutions

- anonymous

I have a test in 4days time on this stuff, double/triple integrals etc

- anonymous

Oh, okay, thanks. Good luck to you! I'll get back here if I can't do it lol.

- anonymous

let A be the angle

- anonymous

x=rcos(A)
y= rsin(A)

- anonymous

the jacobian of the substitution is "f"

- anonymous

edit "r"

- anonymous

how would I get r? is it = 1? I figured out theta, which is from 0 to pi/4

- anonymous

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- anonymous

not the best picture lol

- anonymous

lol, i'll upload mine

- anonymous

the angle goes from 90degrees to 45degrees actually

- anonymous

but the difference is still the same

- anonymous

and our r goes from 0<=r<=1

- 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

This is what I came up with, is it correct?

##### 1 Attachment

- anonymous

Elec, your drawing is correct, but the sector described is actually the portion you have marked 45 degrees. Good job

- 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

wait, the region I drew was correct I am fairly sure

- 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

is it 2y+2x

- anonymous

\[= \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta \int\limits_{0}^{1} r^2 dr \]

- anonymous

= \[\frac{\pi}{4} \times \frac{1}{3} = \frac{\pi}{12}\]

- anonymous

What about the region I drew? Is that acceptable?

- 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

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

But isn't it, 0<=y<=x?

- anonymous

no!

- anonymous

the region you drew was y<=x and y<=sqrt(1-x^2)

- anonymous

the intersection of those region , thats what your picture shows

- 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

Out of curiosity, revan, where did you get sq rt 2/2 from?

- 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

Part of confusion might be: that's not the question you posted here for this post.

- anonymous

Lol, sorry >.< The question I posted is the single iterated integral. I just needed help solving it.

- anonymous

yes, the way you did was very wrong lols

- anonymous

:D thats why i posted lol

- anonymous

\[\int\limits(ax+b)^n = \frac{(ax+b)^{n+1} } {a(n+1)} +C \]

- anonymous

only works for linear functions

- anonymous

raised to a power

- anonymous

Now elec, is writing wrong thing

- anonymous

You said to do poloar coordinates. So, I think I can solve it now. I'll post it when I'm done.

- anonymous

Btw, y is r from 0 to 1? o.0

- anonymous

nvm . ..

- anonymous

Revan, if given this from scratch, you choose polar, but if given the integrals already compiled it is very easy to read.

- anonymous

Yea, but I cant' solve it w/o using polar, but either system, the answer will be the same?

- anonymous

What do you mean by 'solve'

- anonymous

or find volume? that's what you mean?

- 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

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

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

Lol, okay, thx., yea I didn't know wat was going on.

- 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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