Question

How do you find the volume of the solid bounded by x^2+y^2=4 and z=x+y in the first octant?

Answer 1

16/3

Explanation:

you'll want to use

`int int z(x,y) dA`

`= int_(y = 0)^(2) int_(x=0)^(sqrt(4-y^2)) (x+y) \ dx \ dy`

`=int_(y = 0)^(2) [x^2/2+ xy]_(x=0)^(sqrt(4-y^2)) \ dy`

`=int_(y = 0)^(2) 2-y^2/2+ ysqrt(4-y^2) \ dy`

`=[ 2y-y^3/6 - 1/3 (4-y^2)^(3/2) ]_(y = 0)^(2)`

`= 16/3`