Answer to Question #216892 in Calculus for Gayathri Pranathi

Question

Evaluate 3 - 4 x DV where E is the region below z= 4 -x y and above the region in the plane xy plane and defined by 0 < X < 2 and 0< y<1

Answer 1

Solution

$I=\iiint_{V}{(3-4x)dV}=\int_{0}^{1}\int_{0}^{2}\int_{0}^{4-xy}{(3-4x)dz\ dx\ dy}$

$I=\int_{0}^{1}\int_{0}^{2}{(3-4x)(4-xy)dx\ dy}=\int_{0}^{1}\int_{0}^{2}{(12-16x-3xy+4x^2y)dx\ dy}$

$\int_{0}^{2}{(12-16x-3xy+4x^2y)dx}=\left.\ \left(12x-8x^2-\frac{3}{2}x^2y+\frac{4}{3}x^3y\right)\right|_{x=0}^{x=2}=24-32-6y+\frac{32}{3}y=\frac{14}{3}y-8$

$I=\int_{0}^{1}\left(\frac{14}{3}y-8\right)dy=\left.\ \left(\frac{7}{3}y^2-8y\right)\right|_{y=0}^{y=1}=\frac{7}{3}-8=-\frac{17}{3}\approx-5.66667$

Answer

I = -17/3 ≈ -5.66667