Question

Evaluate triple integral ∫ ∫ ∫ 8xydV, where E lies under the plane z = 1+x+y and above the E region in the xy-plane bounded by the curves y = √ x, y = 0, and x = 1.

Answer 1

Answer:

$\mathbf{=\dfrac{163.384}{15}}$

Step-by-step explanation:

$\int \int \limits_{E} \int \ 8 xy dV = \int\limits^{1}_{0} \int\limits^{\sqrt{x}}_{0} \int\limits^{1+x+y}_{0} \ 8xy dz dydx$

$= \int\limits^{1}_{0} \int\limits^{\sqrt{x}}_{0} [ 8xyz]^{z=1+x+y}_{z=0} \ \ dy dx$

$= \int\limits^{1}_{0} \int\limits^{\sqrt{x}}_{0} 8xy (1+x+y) dy dx$

$= \int\limits^{1}_{0} \int\limits^{\sqrt{x}}_{0} 8xy+8x^2y+8xy^2 \ \ dy dx$

$= \int\limits^{1}_{0} \ [ 4xy^2+4x^2y^2+2.7xy^3]^{ y= \sqrt{x}}_{y-0} \ \ dx$

$= \int\limits^{1}_{0} \ 4x (\sqrt{x})^2+4x^2(\sqrt{x})^2+2.7x(\sqrt{x})^3\ \ dx$

$= \int\limits^{1}_{0} \ 4x^2+4x^3+2.7x^{5/2} \ dx$

$\mathbf{=\dfrac{163.384}{15}}$