Find the volume under the surface z = f(x,y) and above the rectangle with the given boundaries.
-$$z = 6 x ^ { 2 } y ; 0 \leq x \leq 4,0 \leq y \leq 3$$
A) 576
B) 1256
C) 2256
D) 676

Question

Quizplus · Accepted Answer

The volume under the surface $z = 6x^2y$ and above the rectangle defined by $0 \leq x \leq 4, 0 \leq y \leq 3$ can be found by integrating the function over the given rectangle. The double integral for the volume $V$ is given by:$$V = \int_{0}^{3} \int_{0}^{4} 6x^2y \, dx \, dy$$First, integrate with respect to $x$:$$\int_{0}^{4} 6x^2y \, dx = 2y[x^3]_{0}^{4} = 2y(64) = 128y$$Then, integrate the result with respect to $y$:$$\int_{0}^{3} 128y \, dy = 128[\frac{1}{2}y^2]_{0}^{3} = 64(9) = 576$$Therefore, the volume under the given surface and above the specified rectangle is 576.

Find the Volume Under the Surface Z = F(x,y) and Above