Solve Polar Integrals - Math Help

[sarahisme]

http://img218.imageshack.us/img218/6003/picture2mz7.png

 

[FrogPad]

You could evaluate them to check your answer.

 

[quantumdude]

Yes, please show the working. If you don't then someone else has to do the problem to check your work, and that isn't the idea of the Homework Help section of the site.

 

[sarahisme]

yep, ok sure thing guys. i'll post my working in a sec... :)

 

[sarahisme]

its part (b) which i am really struggling with...

http://img241.imageshack.us/img241/7764/picture3qq0.png

 

[Astronuc]

a. is correct, and the integrands in the others appear correct, but it would be useful to know how the limits were determined. Please do show your work. Thanks.

 

[0rthodontist]

Your revised working of b. looks correct. In c., the limits of integration are not correct.

 

[sarahisme]

Your revised working of b. looks correct.

ok, but can the radius really go from -1/cos(theta) to 1/cos(theta)? (i.e. how can radius be negative?

 

[sarahisme]

as for (c), here is my working...

http://img215.imageshack.us/img215/903/picture6ax5.png

 

[0rthodontist]

ok, but can the radius really go from -1/cos(theta) to 1/cos(theta)? (i.e. how can radius be negative?

You're right--I was wrong. The region includes the area where theta goes from -pi/4 to pi/4 and also the area where theta goes from 3pi/4 to 5pi/4. You should really just split the two parts of the region up into two integrals, which can be condensed into one.

as for (c), here is my working...

http://img215.imageshack.us/img215/903/picture6ax5.png

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theta should only go from -pi/2 to pi/2. If r <= cos theta then cos theta >= 0.

 

[sarahisme]

You're right--I was wrong. The region includes the area where theta goes from -pi/4 to pi/4 and also the area where theta goes from 3pi/4 to 5pi/4. You should really just split the two parts of the region up into two integrals, which can be condensed into one.

would you please be able to show me how to do this, i can't seem to do it...what graph should i be looking at to do this (to be able see the bowtie?)


-=sarah

 

[0rthodontist]

If |y| <= |x| then you should see this as the region that is vertically between the lines y = x and y = -x. If |x| <= 1 this is the vertical stripe going from -1 to 1 on the x axis. The intersection of these regions is the "bowtie." (if you describe it that way then you must know what it is). In the step where you find |tan theta| <= 1, this can also be satisfied when theta is from 3pi/4 to 5pi/4, though personally I'd trust the diagram more.

 

[sarahisme]

ok, i am still a little confused (sorry :()

this is the graph i am looking at in the x-y plane...the regions enclosing the star shapes are the regions we want ot integrate over, right?

http://img201.imageshack.us/my.php?image=picture5sr6.png

 

[0rthodontist]

Not a graph, a region. A graph is like the graph of a function.
http://img201.imageshack.us/img201/2214/bowtiehl4.png
If you have |y| <= |x| then you can break it up into cases:
1. x >= 0. Then -x <= y <= x.
2. x <= 0. Then x <= y <= -x.

 

[sarahisme]

in which case can't we just do this for part (b):

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

 

[0rthodontist]

No, because that only covers the right half of the bowtie.

 

[sarahisme]

for part (c) i get a graph like this (which is y = +/- sqrt(x-x^2)
http://img201.imageshack.us/img201/9586/picture7yv1.th.png

and so we want to integrate

0<=x<=1
0<=y<=sqrt(x-x^2)

i think that's almost right? :S

 

[sarahisme]

No, because that only covers the right half of the bowtie.

yeah, but i put a 2 out the front and so double the area. because the two halfs of the bowtie have the same area?

 

[0rthodontist]

Oh, I didn't see the 2 before. Yes, that's correct, but not because the halves have the same area--because the function inside depends only on the radius, not the angle.

Now that you've drawn the diagram for part c, what can you say about the limits of integration for the angle?

 

[sarahisme]

Now that you've drawn the diagram for part c, what can you say about the limits of integration for the angle?

ok well here is the limits for the region we want to integrate over in that diagram...

0<=x<=1
0<=y<=sqrt(x-x^2)

0<=x<=1 converts to 0<=r<=1/cos(theta) in polar coords

right?

then

0<=y<=sqrt(x-x^2) converts to ... i am not sure, this is where i get stuck

 

[0rthodontist]

You knew from before that r <= cos theta. So cos theta >= 0. What does that tell you about theta?

 

[sarahisme]

You knew from before that r <= cos theta. So cos theta >= 0. What does that tell you about theta?

that theta >= pi/2 ??

 

[sarahisme]

wait, or that cos(theta) has to be postive (or zero) and this is only true in the first and fourth quadrants, which means that you must integrate theta over -pi/2 to pi/2 ??