Question

Can someone help me solve this iterated integral? Tanks.

Answer 1

Here's the equation.

Answer 2

the one marked red?

Answer 3

Yes, xD

Answer 4

use polar

Answer 5

substitutions

Answer 6

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

Answer 7

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

Answer 8

let A be the angle

Answer 9

x=rcos(A)

y= rsin(A)

Answer 10

the jacobian of the substitution is "f"

Answer 11

edit "r"

Answer 12

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

Answer 13

not the best picture lol

Answer 14

lol, i'll upload mine

Answer 15

the angle goes from 90degrees to 45degrees actually

Answer 16

but the difference is still the same

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

I got it based on the 2 iterated integrals below in pic.
ok, thx, i think i can do it now, 0<=r<=1

Answer 22

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

Answer 23

I know it doesnt impact the final answer , but the integral has x as the lower limit, and the semicircle as the upper limit

Answer 24

is it 2y+2x

Answer 25

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

Answer 26

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

Answer 27

What about the region I drew? Is that acceptable?

Answer 28

Oh yeah, you're right, I was looking at it cross ways. Choose some x, and the y goes from x to the semicircle.

Answer 29

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.

Answer 30

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

Answer 31

no!

Answer 32

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

Answer 33

the intersection of those region , thats what your picture shows

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

yes, the way you did was very wrong lols

Answer 40

:D thats why i posted lol

Answer 41

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

Answer 42

only works for linear functions

Answer 43

raised to a power

Answer 44

Now elec, is writing wrong thing

Answer 45

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

Answer 46

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

Answer 47

nvm . ..

Answer 48

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

Answer 49

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

Answer 50

What do you mean by 'solve'

Answer 51

or find volume? that's what you mean?

Answer 52

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.

Answer 53

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?

Answer 54

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

Answer 55

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

Answer 56

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.