ok, yep i see.Quote:

Originally Posted byThePerfectHacker

so then if we split the integration into 2 regions, can't we just write this as the answer for part (b)?

http://img201.imageshack.us/img201/1121/picture6yb5.png

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- Aug 23rd 2006, 06:54 PMsarahismeQuote:

Originally Posted by**ThePerfectHacker**

so then if we split the integration into 2 regions, can't we just write this as the answer for part (b)?

http://img201.imageshack.us/img201/1121/picture6yb5.png - Aug 23rd 2006, 07:01 PMThePerfectHackerQuote:

Originally Posted by**sarahisme**

$\displaystyle f(\sqrt{x^2+y^2}$ if replaced by $\displaystyle -x,-y$ becomes, $\displaystyle f(\sqrt{(-x)^2+(-y)^2}=f(\sqrt{x^2+y^2}$ so it is symettrical in the xz plane. Thus, that is valid. - Aug 23rd 2006, 07:08 PMsarahismeQuote:

Originally Posted by**ThePerfectHacker**

http://img201.imageshack.us/img201/4...ure8fr3.th.png

as for part (c), i am still struggling with it,

for part (c) i get a graph like this (which is y = +/- sqrt(x-x^2)

http://img201.imageshack.us/img201/9...ure7yv1.th.png

and so we want to integrate

0<=x<=1

0<=y<=sqrt(x-x^2)

but when i try to convert to polar coordinates i end up in a mess.... :(

i think thats almost right? :S - Aug 23rd 2006, 07:30 PMThePerfectHackerQuote:

Originally Posted by**sarahisme**

(I am just cautioning you, I do not know all the specific rules for manipulating polar coordinates, I might be wrong). For example, when radius can be negative, when it cannot, .... Hate multivarible calculus, so informal :mad: )

I do not understand your second problem, rephrase.

as for part (c), i am still struggling with it,

for part (c) i get a graph like this (which is y = +/- sqrt(x-x^2)

http://img201.imageshack.us/img201/9...ure7yv1.th.png

and so we want to integrate

0<=x<=1

0<=y<=sqrt(x-x^2)

but when i try to convert to polar coordinates i end up in a mess.... :(

i think thats almost right? :S[/QUOTE] - Aug 23rd 2006, 07:42 PMsarahisme
ok rephrase, umm... i drew this graph http://img201.imageshack.us/img201/9...ure7yv1.th.png

which is a graph of y = +/- (x-x^2)^(1/2)

now i want to integrate over that circle, right? so then i want to integrate over

0<=x<=1

0<=y<=sqrt(x-x^2)

then i want to convert to polar coordinates.

coverting 0<=x<=1 is easy enough, it just gives you 0<=r<=1/cos(theta)

but i can't seem to convert 0<=y<=sqrt(x-x^2) to polar coordinates - Aug 23rd 2006, 07:43 PMThePerfectHackerQuote:

Originally Posted by**sarahisme**

The most efficient way to solve this problem is with a Jacobain substitution. Since the circle is off center you can place it into the center (origon) by a basic substition of x for u+1, that will move the coordinate axes to the left 1 unit and substitute y for v because you are not changing anything.

Doing it your way.

$\displaystyle x^2+y^2=x$

Thus,

$\displaystyle r^2=r\cos \theta$

Thus,

$\displaystyle r=\cos \theta$

Finished. I looks rather simple in polar form. - Aug 24th 2006, 01:12 AMsarahisme
ok i think i almost have it now....

for part (b) i get:

http://img201.imageshack.us/img201/1121/picture6yb5.png

for (c) i get this answer:

$\displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{cos \theta}}f(tan\theta)rdrd\theta$