Question

How do I compute the double integral over S of x+y (posted in equation editor) where S is the region in the first quadrant (x>0, y>0) inside of the disk x^2 + y^2 <= a^2 and under the line y=sqrt(3)x.

Answer 1

I figure that one should convert to polar form integrating:
\[\int\limits_{0}^{\pi/2}\int\limits_{0}^{a/2}r^2(\cos\theta + \sin\theta) dr d\theta\]

I got a/2 from substituting y=sqrt(3)x into x^2+y^2=a^2

Answer 2

what is sqrt(3)x supposed to be? sqrt(3x)? cube root of 3x?

Answer 3

cube root of 3 multiplied by x

Answer 4

cube root of 3 times x? (3)^(1/3) * x?

Answer 5

sorry square root of 3 so: \[\sqrt{3} \times x\]

Answer 6

well theta isn't from 0 to pi/2

Answer 7

it should be from 0 to pi/3. and
radius is from 0 to a

Answer 8

Right, I understand now that theta isn't 0 to pi/2 but how did you arrive at pi/3?

Answer 9

tan(theta) = y/x = sqrt(3)x/x = sqrt(3)

tan(pi/3) = sqrt(3)

Answer 10

Thanks, didn't think of that!