Findz dV, where E is the solid tetrahedron with vertices (0,0,0), (1,0,0),(0,4,0), and (0,0,2)

Answered: Find z dV, where E is the solid…

Elementary Geometry for College Students

6th Edition

ISBN:9781285195698

Author:Daniel C. Alexander, Geralyn M. Koeberlein

Publisher:Daniel C. Alexander, Geralyn M. Koeberlein

Chapter9: Surfaces And Solids

Section9.CR: Review Exercises

Problem 26CR

See similar textbooks

Related questions

Question

Find
z dV, where E is the solid tetrahedron with vertices (0,0,0), (1,0,0),
(0,4,0), and (0,0,2)

Expert Solution

Step 1

The vertices of the tetrahedron are (0,0,0),(1,0,0),(0,4,0),(0,0,2).

$x varies from x : 0 \to 1 y varies from y : 0 \to 4 z varies from z : 0 \to 2$

The region can be written as E = {(x, y, z)| 0 ≤ x ≤ 1, 0 ≤ y ≤ 4 −4x, 0 ≤ z≤ 2 − 2x −0.5y}.

$\int \int \int_{E} z d v = \int_{0}^{1} \int_{0}^{4 - 4 x} \int_{0}^{2 - 2 x - 0.5 y} (z) d z d y d x = \int_{0}^{1} \int_{0}^{4 - 4 x} {\frac{z^{2}}{2}}_{0}^{2 - 2 x - 0.5 y} d y d x = \frac{1}{2} \int_{0}^{1} \int_{0}^{4 - 4 x} {(2 - 2 x - 0.5 y)}^{2} d y d x = \frac{1}{2} \int_{0}^{1} {\frac{{(y + 4 x - 4)}^{3}}{12}}_{0}^{4 - 4 x} d x = \frac{1}{2} \int_{0}^{1} \frac{- 16 {(x - 1)}^{3}}{3} d x = \frac{- 8}{3} \int_{0}^{1} {(x - 1)}^{3} d x = \frac{- 8}{3} {(\frac{{(x - 1)}^{4}}{4})}_{0}^{1} = \frac{- 8}{3} \times \frac{- 1}{4} = \frac{2}{3}$