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    Evaluate the double integral: (x^2 + y^2) ds

    where s is the area of a semi-circle 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 semi-circle 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?
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    anyone?
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    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:
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    your 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
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    (Original post by JoMo1)
    ..
    Doesn't one of the boundaries being y=x^{3/2} (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).
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    (Original post by DFranklin)
    Doesn't one of the boundaries being y=x^{3/2} (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
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    Still somewhat confused. Do you mean 3 \sqrt{x} (which seems to be what you've just written), or x\sqrt{3}, which seems a bit more plausible given your original post?
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    (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:
    Show
    your 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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    (Original post by DFranklin)
    Doesn't one of the boundaries being y=x^{3/2} (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 probably shoulda read the question...

    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:
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    r=tan^2(theta)sec(theta) in the annoying region, r=1 elsewhere


    although chances are that's wrong because I'm a moron.
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    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?
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    (Original post by DFranklin)
    Still somewhat confused. Do you mean 3 \sqrt{x} (which seems to be what you've just written), or x\sqrt{3}, which seems a bit more plausible given your original post?

    x\sqrt{3}

    I mean this one sorry! I've never really learnt latex etc. know I should, sorry!
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    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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    (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?
    Ok, if it's a pizza slice then you need to find the angle that your 2 boundary lines follow, then those are your theta boundary conditions. e.g. if your boundaries went through the middle of the top 2 sectors then your boundary conditions would be Pi/2 and 3Pi/2.

    and having written that mess of *******s I'm going to go off and learn LateX. This also forms good procrastination.
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    (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?
    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...
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    (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...
    You know the gradient of your boundary lines because you have their equations, what does the gradient represent in practical terms? It might help to draw some diagrams making your boundary lines into the hypotenuse for a right-angled triangles.

    Edit: and remember polar co-ordinates are measured anti-clockwise from the principal-axis (the positive x-axis), not from which ever axis is closest
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    So your integral is

    \int_{S} x^2 + y^2 dS = \int_{0}^{2a} r^2 dr \int_{A} d\theta

    where A is between the angles the line segments y=-x and y=sqrt(3)x above the x-axis 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.
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    (Original post by StandardCarpet)
    So your integral is

    \int_{S} x^2 + y^2 dS = \int_{0}^{2a} r^2 dr \int_{A} d\theta

    where A is between the angles the line segments y=-x and y=sqrt(3)x above the x-axis 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!!!
    ahhh right, yeah I can see how the upper limit is 3pi/4, as it cuts the section in half, but I don't know how you can .....

    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
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    (Original post by StandardCarpet)
    So your integral is

    \int_{S} x^2 + y^2 dS = \int_{0}^{2a} r^2 dr \int_{A} d\theta

    where A is between the angles the line segments y=-x and y=sqrt(3)x above the x-axis 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.
    Yeah, I'm having a mare with multiple integrals plus I hate grout's lecture notes!
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    (Original post by danhirons)
    Yeah, I'm having a mare with multiple integrals plus I hate grout's lecture notes!
    Grout's "lecture notes"? You mean ramblings, "old friend", "nephew who figured out probability on my knee" and all that crap!
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    (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!
    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
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    (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 got (pi*a^4)/3.

    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 y-axis, 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!!
 
 
 
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