Question

, where \({\rm{E}}\)is bounded by the parabolic cylinders \({\rm{y = }}{{\rm{x}}^{\rm{2}}}\) and \({\rm{x = }}{{\rm{y}}^{\rm{2}}}\)and the planes \({\rm{z = 0}}\)and \({\rm{z = x + y}}\).

Short answer

The required value is\(\frac{{\rm{3}}}{{{\rm{28}}}}\).

Step-by-step solution

Concept Introduction

Triple integrals are the three-dimensional equivalents of double integrals. They're a way to add up an unlimited number of minuscule quantities connected with points in a three-dimensional space.

If the area under the curve from to is the definite integral of a function of one variable, then the double integral is equal to the volume under the surface and above the -plane in the integration region.

Find the value of.

The solid is defined above the region defined by two parabolas intersecting at the point\({\rm{(1,1)}}\)defined by the equations\({\rm{x = }}{{\rm{y}}^{\rm{2}}}\)and\({\rm{y = }}{{\rm{x}}^{\rm{2}}}\). The following triple integral is obtained by expressing y from the first equation:

\(\int_{\rm{0}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\sqrt {\rm{x}} } {\int_{\rm{0}}^{{\rm{x + y}}} {\rm{x}} } } {\rm{ydzdydx = }}\int_{\rm{0}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\sqrt {\rm{x}} } {{{\rm{x}}^{\rm{2}}}} } {\rm{y + x}}{{\rm{y}}^{\rm{2}}}{\rm{dydx}}\)

The remaining double integral is then calculated:

\(\begin{array}{c}\int_{\rm{0}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\sqrt {\rm{x}} } {{{\rm{x}}^{\rm{2}}}} } {\rm{y + x}}{{\rm{y}}^{\rm{2}}}{\rm{dydx = }}\int_{\rm{0}}^{\rm{1}} {\left( {\frac{{{\rm{3}}{{\rm{x}}^{\rm{3}}}{\rm{ + 2}}\sqrt {{{\rm{x}}^{\rm{5}}}} {\rm{ - 3}}{{\rm{x}}^{\rm{6}}}{\rm{ - 2}}{{\rm{x}}^{\rm{7}}}}}{{\rm{6}}}} \right)} {\rm{dx}}\\{\rm{ = }}\frac{{\rm{3}}}{{{\rm{28}}}}\end{array}\)

Therefore, the required solution is\(\frac{{\rm{3}}}{{{\rm{28}}}}\).