Calculus 3 double integrals

Question

Find the volume of the region under the surface z=xy^2 and above the area bounded by x=y^2 and x-y=2

Answer 1

Draw the region of integration on the xy-plane.

Find the points of intersection of the parabola x = y² with the line x = y + 2, being ( 1, -1) and (4,2 ).

Then the volume is

∫_y=-1^y=2 ∫^x=y+2_x=y^2 (xy²⁾ dxdy = (1/2) ∫_y=-1^y=2[x²y²]^x=y+2_x=y^2dy = (1/2) ∫_y=-1^y=2[ (y+2)²y²-y⁴]dy

=(1/2) ∫_y=-1^y=2 [ y⁴+4y³+4y²-y⁴]dy =(1/2) ∫_y=-1^y=2[ 4y³ +4y²] dy =(1/2) [ y⁴ + (4/3) y³]-₁² =

(1/2) [ 16 +(32/3) -1+(4/3)] = (1/2)[15 +(36/3)] =27/2 cubic units

Answer 2

x = y² and x - y = 2

x = y + 2; y + 2 = y²; y² - y - 2 = 0; (y + 1)(y - 2 ) = 0; y = - 1, y = 2.

V = ∫∫_Rzdxdy; R: -1 ≤ y ≤ 2, y² ≤ x ≤ y + 2

V = ∫_-1²∫_y^2^y+2xy²dxdy∫ = ∫_-1²y²x²/2_y^2^y+2dy = 1/2∫_-1²y²(y² + 4y + 4 - y⁴)dy = 1/2∫_-1²(y⁴ + 4y³ + 4y² - y⁶)dy =

1/2(y⁵/5 + y⁴ + 4y³//3 - y⁷/7)_-1² = 1/2[(32/5 + 16 + 32/3 - 128/7) - (- 1/5 + 1 - 4/3 + 1/7)] =

1/2(33/5 +15 + 12 - 129/7) = 531/70 = 7.5857