Question

where\({\rm{T}}\)is the solid tetrahedron with vertices\({\rm{(0,0,0),(1,0,0),(0,1,0), and (0,0,1)}}\).

Short answer

The required answer is\(\frac{{\rm{1}}}{{{\rm{60}}}}\).

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 trick is to find at least one projection of the supplied plane onto one of the planes indicated by the coordinate axes in order to find the region of integration. We can then establish a boundary on the second integral after that. One such projection can be obtained by setting\({\rm{z = 0}}\). We arrive at the following conclusion:

\(\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - x}}} {\int_{\rm{0}}^{{\rm{1 - x - y}}} {{{\rm{x}}^{\rm{2}}}} } } {\rm{dzdydx = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - x}}} {{{\rm{x}}^{\rm{2}}}} } {\rm{ - }}{{\rm{x}}^{\rm{3}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{ydydx}}\)

Calculate the remainder of the double integral:

\(\begin{array}{c}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - x}}} {{{\rm{x}}^{\rm{2}}}} } {\rm{ - }}{{\rm{x}}^{\rm{3}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{ydydx = }}\int_{\rm{0}}^{\rm{1}} {\frac{{{{\rm{x}}^{\rm{4}}}{\rm{ - 2}}{{\rm{x}}^{\rm{3}}}{\rm{ + }}{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} {\rm{dx}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{60}}}}\end{array}\)

Therefore, the required solution is\(\frac{{\rm{1}}}{{{\rm{60}}}}\).