Question

`intint_R(x+y)dA` by circle area is `x^2+y^2=9`?

`intint_R(x+y)dA` When R is a place surrounded by Circle is`x^2+y^2=9`

Answer 1

= 0

Explanation:

`int_A x + y \ dA`

`= int_(-3)^(3) dx \ int_(-sqrt(9 -x^2))^(sqrt(9 -x^2)) dy qquad x + y`

`= int_(-3)^(3) dx qquad [xy + y^2/2]_( - sqrt(9 -x^2))^( sqrt(9 -x^2))`

`= int_(-3)^(3) dx qquad 2xsqrt(9 -x^2)`

`= int_(-3)^(3) dx qquad d/(dx) [- 2/3(9 -x^2)^(3/2) ]`

`= [- 2/3(9 -x^2)^(3/2) ]_(-3)^(3) bb(= 0)`

From the symmetry, what you would expect

In polar:

`= int_0^(2 pi) d theta int_0^3 dr qquad r * ( r cos theta + r sin theta )`

`= int_0^(2 pi) d theta qquad [r^3/3 ( cos theta + sin theta )]_0^3`

`= 9 int_0^(2 pi) d theta qquad cos theta + sin theta`

`= 9 [ sin theta - cos theta ]_0^(2 pi)`

`= 9( [ 0 - 1 ] - [0 - 1] ) = 0`