Answer to Question #187008 in Calculus for Jethro

Question #187008

Use a triple Integral to determine the volume of the tetrahedron with vertices (0,0,0), (3,0,0), (0,4,0) and (0,0,5).


1
Expert's answer
2021-05-07T12:36:28-0400

The equation of the line that passes throughout the points (3,0,0) and (0,4,0) in the plane xOyxOy is x3+y4=1\frac{x}{3}+\frac{y}{4}=1 and hence y=4(1x3)y=4(1-\frac{x}{3}). The equation of the plane that passes throughout the points (3,0,0), (0,4,0) and (0,0,5) is x3+y4+z5=1\frac{x}{3}+\frac{y}{4}+\frac{z}{5}=1 and thus z=5(1x3y4)z=5(1-\frac{x}{3}-\frac{y}{4}). It follows that the volume VV is uqual to


V=03dx04(1x3)dy05(1x3y4)dz=03dx04(1x3)5(1x3y4)dy=503(4(1x3)4x3(1x3)18(4(1x3))2)dx=503(44x34x3+4x292+4x3+2x29)dx=503(24x3+2x23)dx=5(2x2x23+2x39)03=5(66+2)=10V=\int_{0}^{3}dx\int_{0}^{4(1-\frac{x}{3})}dy\int_0^{5(1-\frac{x}{3}-\frac{y}{4})}dz= \int_{0}^{3}dx\int_{0}^{4(1-\frac{x}{3})}5(1-\frac{x}{3}-\frac{y}{4})dy= 5\int_{0}^{3}(4(1-\frac{x}{3})-4\frac{x}{3}(1-\frac{x}{3})-\frac{1}{8}(4(1-\frac{x}{3}))^2)dx= 5\int_{0}^{3}(4-\frac{4x}{3}-4\frac{x}{3}+\frac{4x^2}{9}-2+4\frac{x}{3}+2\frac{x^2}{9})dx= 5\int_{0}^{3}(2-\frac{4x}{3}+2\frac{x^2}{3})dx= 5(2x-\frac{2x^2}{3}+\frac{2x^3}{9})|_0^3=5(6-6+2)=10


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