Converting Cartesian to polar: semicircle or full circle bounds

[asdf12312]

Homework Statement

Homework Equations

The Attempt at a Solution

my only problem curently is in finding the angle θ. I do get the equation x^2 + y^2 =1 however am confused whether this would be a semi-circle on the positive axis or a full circle. because my teacher has notes that confuse me. for instance ∫^{$\sqrt{1-y^2}$} _{-$\sqrt{1-y^2}$} my formatting is a bit off but that would be higher/lower bound. that is the only case where she made a complete circle, instead of a semi circle.

if it is a semi circle, theta would be 0≤ θ ≤ PI/2
however if its a full circle, since from y=-1 to y=1 it would be -PI/2 ≤ θ≤ PI/2.

 

[LCKurtz]

x goes from 0 to the positive square root of $1-y^2$. What part of the circle do you get when you solve it for $x$ and take the positive root? That will tell you which $\theta$'s to use.

 

[asdf12312]

i think it would be the right hemisphere of the circle, since x=0 and increases, and the circle has radius of 1. so -PI/2 ≤ θ ≤ PI/2 what i got is right?

 

[LCKurtz]

i think it would be the right hemisphere of the circle, since x=0 and increases, and the circle has radius of 1. so -PI/2 ≤ θ ≤ PI/2 what i got is right?

Yes.

 

[asdf12312]

OK. then i got r³ by looking at f(x,y) so I ended up integrating ∫(r⁴ dr)dθ and got my final answer as PI/5.

 

[HallsofIvy]

Integrating with r from 0 to 1, $\theta$ from $-\pi/2$ to $\pi/2$.

Yes, that is correct.