# Calculus questions

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• Aug 23rd 2006, 06:54 PM
sarahisme
Quote:

Originally Posted by ThePerfectHacker
No,
$\displaystyle |x|\geq |y|$
$\displaystyle |x|\leq 1$

ok, yep i see.

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 PM
ThePerfectHacker
Quote:

Originally Posted by sarahisme
ok, yep i see.

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

You are using symettry, good. But there is one thing which you might not have considered. The regions are symettrical, yes. But what about the surface above it! It happens to be true since,
$\displaystyle f(\sqrt{x^2+y^2}$ 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 PM
sarahisme
Quote:

Originally Posted by ThePerfectHacker
You are using symettry, good. But there is one thing which you might not have considered. The regions are symettrical, yes. But what about the surface above it! It happens to be true since,
$\displaystyle f(\sqrt{x^2+y^2}$ 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.

ok, sweet. if the surface above wasn't symmetric, would the answer be this for (b) :

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 PM
ThePerfectHacker
Quote:

Originally Posted by sarahisme
ok, sweet. if the surface above wasn't symmetric, would the answer be this for (b) :

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

I assume so.

(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 PM
sarahisme
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 PM
ThePerfectHacker
Quote:

Originally Posted by sarahisme

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

I understand you now.
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 AM
sarahisme
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:

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