Calculus 3 Triple Integrals

Question

Find ∫∫∫E zdV, where E is the solid tetrahedron with vertices (0,0,0), (5,0,0), (0,1,0), and (0,0,2)

Answer 1

The solid over which we are going to integrate is on the first octant bounded by the plane

x = 0, y = 0, z = 0 and the plane x/5 +y + z/2 =1.

So ∫^x=5_x=0 ∫^{y= 1-x/5}_y=0 ∫^{z= 2-2x/5-2y}_z=0[ z] d z d y d x =

(1/2) ∫^x=5_x=0 ∫^{y= 1-x/5}_y=0[z]²|^z=2-2x/5-2y_z=0 d y d x =

(1/2) ∫^x=5_x=0 ∫^{y= 1-x/5}_y=0[ 2−2x/5−2y]² d y d x=

(-1/12) ∫^x=5_x=0[ 2−2x/5−2y]³|^1-x/5₀ dx=

(-1/12) ∫^x=5_x=0{[2-2x/5−2(1-x/5)]³ −[2−2x/5]³} dx=

(-1/12) ∫^x=5_x=0[2x/5−2]³dx=

(-5/24) ∫^w=0_w=-2w³dw=

(-5/96) [w⁴]|⁰_w=-2=

5/6