- #1
Xyius
- 508
- 4
Hello everybody, I am having trouble doing this polar double integral. The problem says..
Find the area of the region..
[tex]\frac{1}{2}y^2 \leq x \leq 2y
0 \leq y \leq 8[/tex]
It is hard for me to come up with the limits of integration. Checking the answer would be easy because I can compare it to the rectangular integral which is easy, but I keep getting it wrong! The most trouble I am having is converting the line to a polar form and using it in the integral. I am pretty sure it would be theta = arctan(2) but I do not know where to put this in my integral.
For the upper limit on "r" I have [tex]4cot(\theta)csc(\theta)[/tex] I do not know what to put as the lower limit on r except the length r is at that point on the graph of [tex]4cot(\theta)csc(\theta)[/tex]. Which I got to be [tex]\frac{2}{sin(arctan(2)}[/tex]
I am very tired at the moment (Its 1:30 am!) and it is likely this doesn't make any sense to you. I just need some help. :\
Find the area of the region..
[tex]\frac{1}{2}y^2 \leq x \leq 2y
0 \leq y \leq 8[/tex]
It is hard for me to come up with the limits of integration. Checking the answer would be easy because I can compare it to the rectangular integral which is easy, but I keep getting it wrong! The most trouble I am having is converting the line to a polar form and using it in the integral. I am pretty sure it would be theta = arctan(2) but I do not know where to put this in my integral.
For the upper limit on "r" I have [tex]4cot(\theta)csc(\theta)[/tex] I do not know what to put as the lower limit on r except the length r is at that point on the graph of [tex]4cot(\theta)csc(\theta)[/tex]. Which I got to be [tex]\frac{2}{sin(arctan(2)}[/tex]
I am very tired at the moment (Its 1:30 am!) and it is likely this doesn't make any sense to you. I just need some help. :\