Question

Double integral, problem in comments. :)

Answer 1

\[\int\limits_{}^{}\int\limits_{}^{}_{D}\frac{ 1 }{ 4 }x(\frac{ 1 }{ 2 }+y)dxdy\]
Where D is the limited area x>0, enclosed by x^2+y^2=32 and y^2=4x

Answer 2

I'm having a hard time getting good domains

Answer 3

according to me
x will go from 0 to 4 and y will go from 0 to 4

Answer 4

Why?

Answer 5

sorry y will go 0 to 2sqrt(x)

Answer 6

i found out the point of intersection
by the use of equation of the circle given and equation of parabola
it is x^2 + y ^2 = 32 and y^2 = 4x
now x^2 +4x = 32
x^2 +8x -4x -32 = 0
after solving x= -8 and x= 4
x=-8 rejected as value of x>0 is given in ques

Answer 7

no
x(x+8) -4(x+8) = 0
(x-4)(x+8) = 0
x-4 =0 or x+8 = 0
x = 4 or x = -8