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Multiple Integrals

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Avatar for danhirons
danhirons
OP
15
Original post by StandardCarpet
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!!


i only copied the limit from you, you sure about that? grad will be y/x, which will be the same as tan = o/a for the angle between the line and the x axis. Surely even if the angle with the +ve y axis was pi/3 then angle with the positive x axis would pi/2 - pi/3 anyway?

I'm pretty tired as I'm writing this so I may be completely messing things up....
Original post by danhirons
i only copied the limit from you, you sure about that? grad will be y/x, which will be the same as tan = o/a for the angle between the line and the x axis. Surely even if the angle with the +ve y axis was pi/3 then angle with the positive x axis would pi/2 - pi/3 anyway?

I'm pretty tired as I'm writing this so I may be completely messing things up....


Well think about it this way: sqrt(3) is greater than one, so the line segment will be steeper than a line of y=x. We agree on at least one of the angles - either the one between the line and the positive y-axis, or the one between the line and the positive x-axis, which is the one we want - being arctan(sqrt(3)) = pi/3. Therefore the other angle must be 2pi/3.

Since the gradient of the line is greater than one, the angle the line makes with the positive x-axis - the one we want - must be greater than the angle the line makes with the positive y-axis. Therefore the angle we want must be 2pi/3.

I think!
Reply 22
Avatar for danhirons
danhirons
OP
15
Original post by StandardCarpet
Well think about it this way: sqrt(3) is greater than one, so the line segment will be steeper than a line of y=x. We agree on at least one of the angles - either the one between the line and the positive y-axis, or the one between the line and the positive x-axis, which is the one we want - being arctan(sqrt(3)) = pi/3. Therefore the other angle must be 2pi/3.

Since the gradient of the line is greater than one, the angle the line makes with the positive x-axis - the one we want - must be greater than the angle the line makes with the positive y-axis. Therefore the angle we want must be 2pi/3.

I think!


I really don't know why I can't visualise what you're saying!!! Well if you are right.......:tongue:

Surely if the smaller angle was 2pi/3 then it would be in the top left quadrant of the set of axis, which it clearly can't be? If we're still confused after this we'll just have to agree to disagree :biggrin:
Original post by danhirons
I really don't know why I can't visualise what you're saying!!! Well if you are right.......:tongue:

Surely if the smaller angle was 2pi/3 then it would be in the top left quadrant of the set of axis, which it clearly can't be? If we're still confused after this we'll just have to agree to disagree :biggrin:




tan(theta) = Opposite / Adjacent = dx/dy = 1/sqrt(3)

theta = arctan(1/sqrt(3)) = pi/6

Therefore, alpha = pi/2 - pi/6 = pi/3

Right, lower limit must be pi/3!

I'm sorry, above I was somehow set in thinking that 90 degrees was pi, rather than pi/2.
(edited 13 years ago)
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