Double integral(D) ydA, where D is the region in the first quadrant bounded by the parabolas x=y^2 and x=8-y^2

Question

Calculus 3 Math. I need help solving this.

Answer 1

Obviously the curves will intersect when y^2 = 4, i.e. y = ±2; but we are restricted to the first quadrant, so we only use y = 2. In this region (y = 0 to 2) the higher boundary for x is 8-y^2 and the lower boundary is y^2. So we have
∫(0 to 2) ∫(y^2 to 8-y^2) y dx dy
= ∫(0 to 2) y (8-y^2 - y^2) dy
= ∫(0 to 2) (8y - 2y^3) dy
= [4y^2 - y^4/2][0 to 2]
= (4(4) - 16/2) - (4(0) - 0/2)
= 8.

Answer 2

6

Answer 3

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Answer 4

Wow.

I'll do my best:
x=pi.

Answer 5

do your own homework!!