Polar double integrals: Finding the angle boundaries

[mg236]

Hey so I'm having trouble finding the correct angles for certain problems. I've got two examples. Don't actually need to do the integral, just trying to understand how to find the right boundaries for theta.

1)Finding the mass of the surface which is the hemisphere z = squareroot(4-x^2-y^2) and the density function p(x,y,z) = |xy|
So here as we have a circle in the xy plane I would think that the angle goes from 0 to 2pi but instead it seems like it goes from 0 to pi/2 and we have to multiply the integral by 4. Why doesn't 0 to 2pi work?
2) Finding the surface area of the portions of the sphere x^2+y^2+z^2 = a^2 that are within the cylinder x^2+y^2 = ay
This time we have a circle that can be written as r = asin(theta). My initial instinct would be to calculate the angle from 0 to pi as the circle is right above the x axis, but it seems that once again the correct angles are from 0 to pi/2.. Again, why is 0 to pi wrong?

Just feels like I'm missing something. Most problems I can do fine but some problems I just plug in the wrong angles and I'm not sure why the angles I'm using are wrong. Like there's something that I don't quite understand yet. Can anyone help?

 

[dunkelheit]

1) [MATH]\theta \in [0,2\pi)[/MATH] works fine too, but then you have to split the integral in a sum of integrals because [MATH]|\cos \theta \sin \theta|[/MATH] changes as [MATH]\theta[/MATH] changes in [MATH][0,2\pi)[/MATH].
To avoid this, he is just considering that the integrand is an even function both in [MATH]x[/MATH] and [MATH]y[/MATH] and so, by symmetry, he/she consider only the part with [MATH]x\geq 0[/MATH] and [MATH]y \geq 0[/MATH] and multiplies by [MATH]4[/MATH] the integral; notice that if [MATH]x\geq 0[/MATH] and [MATH]y \geq 0[/MATH] it follows that [MATH]|xy|=xy[/MATH] and in polar coordinates [MATH]x \geq 0[/MATH] and [MATH]y\geq 0[/MATH] are [MATH]\rho \cos \theta \geq 0[/MATH] and [MATH]\rho \cos \theta \geq 0[/MATH], so by this it follows that [MATH]\cos \theta \geq 0[/MATH] and [MATH]\sin \theta \geq 0[/MATH] and so finally [MATH]\theta \in \left[0,\frac{\pi}{2}\right][/MATH].

 

[Steven G]

It is possible that you are not wrong. Not saying that you are guessing but random or partial guessing does not work. One needs to truly analyze what is going on. Does \(\displaystyle \theta\) really go from 0 to 2pi, or does it go from -pi/2 to pi/2, etc.

Something that is generally helpful is to use symmetry. Have you noticed sometimes that it is easier to evaluate an integral if one of it's limits is 0 instead of -a? I hope that you have. If you have an even function and you want to integrate it from -a to a it might be easier to integrate it from 0 to a and double the answer. Actually the integration would not be easier, as the integration is (usually) independent of the limits, but the evaluation will probably be easier plugging in 0 versus plugging in -a. So in the end, you might go from -a to a and your textbook/instructor goes from 0 to a and neither of you are wrong.

 

[LCKurtz]

Hey so I'm having trouble finding the correct angles for certain problems. I've got two examples. Don't actually need to do the integral, just trying to understand how to find the right boundaries for theta.

2) Finding the surface area of the portions of the sphere x^2+y^2+z^2 = a^2 that are within the cylinder x^2+y^2 = ay
This time we have a circle that can be written as r = asin(theta). My initial instinct would be to calculate the angle from 0 to pi as the circle is right above the x axis, but it seems that once again the correct angles are from 0 to pi/2.. Again, why is 0 to pi wrong?

[MATH]0[/MATH] to [MATH]\pi[/MATH] is not wrong for the polar angle [MATH]\theta[/MATH]. In fact, it is the natural way to approach problem [MATH]2[/MATH].