Evaluate the double integral: (x^2 + y^2) ds
where s is the area of a semicircle x^2 + y^2 = 4a^2, where y>or equal to 0, and defined by the lines y=x and y=x(3)^1/2
My working:
 the domain is a semicircle above the x axis
 best to change it into polar coordinates, x=rcostheta, y=rsintheta, ds= rdrdtheta
 then i just carry out the multiple integral
What I am having problems with is determining what the boundary conditions will be?

danhirons
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 14042011 16:32

danhirons
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 14042011 17:22
anyone?

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 14042011 18:13
Your new cooordinate system is in (r, theta), let's assume you're going to integrate with r then theta, so when you fix r=a, what values of theta can you have?
Now, what values can r vary between?
Spoiler:Showyour r and theta don't depend on one another. Whatever r you have your theta goes from 0..Pi, and your r is 0..1 
DFranklin
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 14042011 18:19
(Original post by JoMo1)
.. 
danhirons
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 14042011 18:24
(Original post by DFranklin)
Doesn't one of the boundaries being (assuming I've interpreted the original post correctly) somewhat invalidate that? (And the y=x line means you can't have theta = pi, although this is easier to work around).
I mean y=3 x sqrt(x). Sorry for any confusion 
DFranklin
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 14042011 18:26
Still somewhat confused. Do you mean (which seems to be what you've just written), or , which seems a bit more plausible given your original post?
Last edited by DFranklin; 14042011 at 18:27. 
danhirons
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 14042011 18:29
(Original post by JoMo1)
Your new cooordinate system is in (r, theta), let's assume you're going to integrate with r then theta, so when you fix r=a, what values of theta can you have?
Now, what values can r vary between?
Spoiler:Showyour r and theta don't depend on one another. Whatever r you have your theta goes from 0..Pi, and your r is 0..1
I'm still rather confused. If I draw the domain of integration isn't the area I'm calculating like an upside down pizza slice ( not sure what the shape is called)? So I can't understand what the boundary conditions would be.
All I can think of is r would be from 0 to 2a and theta would be 0 to the angle of the pizza slice shape (round bottomed triangle sounds better actually) maybe? 
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 14042011 18:31
(Original post by DFranklin)
Doesn't one of the boundaries being (assuming I've interpreted the original post correctly) somewhat invalidate that? (And the y=x line means you can't have theta = pi, although this is easier to work around).
OP: my answers were wrong but the basic principle is the same for finding boundary conditions. However, for this one I'd fix theta and find the associated max r in terms of theta.
I get:
Spoiler:Showr=tan^2(theta)sec(theta) in the annoying region, r=1 elsewhere
although chances are that's wrong because I'm a moron. 
DFranklin
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 14042011 18:31
You've got a pizza slice. So it has two straight edges.
What value of theta corresponds to each of the edges?
So what is the acceptable range of values for theta? 
danhirons
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 14042011 18:31
(Original post by DFranklin)
Still somewhat confused. Do you mean (which seems to be what you've just written), or , which seems a bit more plausible given your original post?
I mean this one sorry! I've never really learnt latex etc. know I should, sorry! 
DFranklin
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 14042011 18:32
OK, in that case, JoMo is right that the range for r doesn't depend on theta, but he's wrong about the acceptable range. See my post #9.

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 14042011 18:33
(Original post by danhirons)
I'm still rather confused. If I draw the domain of integration isn't the area I'm calculating like an upside down pizza slice ( not sure what the shape is called)? So I can't understand what the boundary conditions would be.
All I can think of is r would be from 0 to 2a and theta would be 0 to the angle of the pizza slice shape (round bottomed triangle sounds better actually) maybe?
and having written that mess of *******s I'm going to go off and learn LateX. This also forms good procrastination. 
danhirons
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 14042011 18:41
(Original post by DFranklin)
You've got a pizza slice. So it has two straight edges.
What value of theta corresponds to each of the edges?
So what is the acceptable range of values for theta?
Maybe the lower limit would be pi/4 as it cuts the quadrant in half? Then no idea how to calculate the upper limit angle... 
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 14042011 18:44
(Original post by danhirons)
This is what I don't understand how to calculate!
Maybe the lower limit would be pi/4 as it cuts the quadrant in half? Then no idea how to calculate the upper limit angle...
Edit: and remember polar coordinates are measured anticlockwise from the principalaxis (the positive xaxis), not from which ever axis is closestLast edited by JoMo1; 14042011 at 18:45. 
StandardCarpet
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 14042011 18:53
So your integral is
where A is between the angles the line segments y=x and y=sqrt(3)x above the xaxis make with the positive x axis.
By inspection the upper limit would be 3pi/4, and by tan(theta)=y/x, the lower limit would be arctan(sqrt(3)), which is pi/3.
I think!!!
This isn't off Grout's Fun for Easter, is it? I could check what I got if you give me the question number.Last edited by StandardCarpet; 14042011 at 18:56. 
danhirons
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 14042011 18:58
(Original post by StandardCarpet)
So your integral is
where A is between the angles the line segments y=x and y=sqrt(3)x above the xaxis make with the positive x axis.
By inspection the upper limit would be 3pi/4, and by tan(theta)=y/x, the lower limit would be arctan(sqrt(3)), which is pi/3.
I think!!!
Edit: ahhh thankyou, I can see where the limits come from, suppose I didn't think of the gradient as being y/x ....
Thanks everyone 
danhirons
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 14042011 19:02
(Original post by StandardCarpet)
So your integral is
where A is between the angles the line segments y=x and y=sqrt(3)x above the xaxis make with the positive x axis.
By inspection the upper limit would be 3pi/4, and by tan(theta)=y/x, the lower limit would be arctan(sqrt(3)), which is pi/3.
I think!!!
This isn't off Grout's Fun for Easter, is it? I could check what I got if you give me the question number. 
StandardCarpet
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 14042011 19:05
(Original post by danhirons)
Yeah, I'm having a mare with multiple integrals plus I hate grout's lecture notes! 
danhirons
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 14042011 19:13
(Original post by StandardCarpet)
Grout's "lecture notes"? You mean ramblings, "old friend", "nephew who figured out probability on my knee" and all that crap!
what answer did you get btw? mine simplified out quite nicely at the end surprisingly  which is making me think its wrong.
its q 14b 
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 15042011 13:40
(Original post by danhirons)
Haha yeahhh,
what answer did you get btw? mine simplified out quite nicely at the end surprisingly  which is making me think its wrong.
its q 14b
I actually realised I made a couple of mistakes above; the lower limit for theta is 2pi/3, since pi/3 is the angle made with the positive yaxis, if you look at the trigonometry involved. And also, the integral in r should be r^3, not r^2  sorry, I was doing things off the top of my head with the above; you should always use a sheet of paper for these things!!Last edited by StandardCarpet; 15042011 at 13:43.
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