Find Volume of tetrahedron bounded by the the planes

S

shivers20

Junior Member
Joined
Mar 3, 2006
Messages
68
Find Volume of tetrahedron bounded by the the planes using integrals
Question Details: Find Volume of tetrahedron bounded by the the planes z=0, x=0, y=0 and x+y+z=1.

This is what I have so far,

x+y+z=1

x+y=1

y=1-x

(1-x-y)dy dx

(Integral from 0 to 1 and Integral y= 0 to y=(1-x)

V= (1-x-y) dy dx

(y-xy-y^2/2)]0 to 1-x

((1-x)-x(1-x)-((1-x)^2)/2... dx

(1-x)-(x+x^2)-((x^2-2x+1)... dx

(1-x)-(x+x^2)-((x^2-2x+1)... dx

(x^2-x+(1/2) dx

(x^3/3) + (-x^2/2) + (x^3/3) ] 0 to 1

1/3 -1/2 +1/2 = 1/3

Answer: 1/3
 
shivers20 said:
Find Volume of tetrahedron bounded by the the planes using integrals
Question Details: Find Volume of tetrahedron bounded by the the planes z=0, x=0, y=0 and x+y+z=1.

This is what I have so far,

x+y+z=1

x+y=1

y=1-x

(1-x-y)dy dx

(Integral from 0 to 1 and Integral y= 0 to y=(1-x)

V= (1-x-y) dy dx

(y-xy-y^2/2)]0 to 1-x

((1-x)-x(1-x)-((1-x)^2)/2... dx

(1-x)-(x+x^2)-((x^2-2x+1)... dx <--- Sign problem in the second parenthesis. Also, where did '/2' go?

(1-x)-(x+x^2)-((x^2-2x+1)... dx

(x^2-x+(1/2) dx

(x^3/3) + (-x^2/2) + (x^3/3) ] 0 to 1

1/3 -1/2 +1/2 = 1/3

Answer: 1/3
Click to expand...
 
You must log in or register to reply here.
Top