Find the triple integral?

Question

Find the triple integral over the region of x^2 + y + z dV if the region R in space is bounded by the planes x=0, y=0, z=0, and x + y + z = 1

Answer 1

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0 ≤ x ≤ 1 ∫ 0 ≤ y ≤ 1 - x ∫ 0 ≤ z ≤ 1 - y - x ∫ ( x^2 + y + z ) dz dy dx

1 - x - y

∫ ( x^2 + y + z ) dz

0

................... . . . . . . . . . .1 - y - x

( x^2 z + y z + (1/2) * z^2 ) ]

................... . . . . . . . . . 0

x^2 * ( (1 - x - y) - 0 ) + y ( (1 - x - y) - 0 ) + (1/2) * ( (1 - x - y)^2 - 0^2 )

x^2 - x^3 - x^2 y + y - xy - y^2 + (1/2) * ( x^2 + 2xy - 2x + y^2 - 2y + 1 )

x^2 - x^3 - x^2 y + y - xy - y^2 + (1/2)x^2 + xy - x + (1/2)y^2 - y + (1/2)

(3/2)x^2 - x^2 y - x^3 - (1/2) y^2 - x + (1/2)

1 - x

∫ ( (3/2)x^2 - x^2 y - x^3 - (1/2) y^2 - x + (1/2) ) dy

0

( (3/2)x^2 y - (1/2) x^2 y^2 - x^3 y - (1/6) y^3 - x y + (1/2) y )---> from 0 to 1 - x

( (3/2)x^2 ( (1 - x) - 0 ) - (1/2) x^2 ( (1-x)^2 - 0^2 ) - x^3 ( (1 - x) - 0 ) - (1/6) ( (1 - x)^3 - 0^3 ) - x ( 1-x - 0 ) + (1/2) ( 1 - x - 0 )

(3/2)x^2 (1 - x) - (1/2) x^2 (1-x)^2 - x^3 (1 - x) - (1/6) (1 - x)^3 - x ( 1 - x ) + (1/2) ( 1 - x )

(3/2)x^2 - (3/2)x^3 - (1/2) x^2 (1 - 2x + x^2) - x^3 + x^4 - (1/6) ( -x^3 + 3x^2 - 3x + 1 ) - x + x^2 + (1/2) - (1/2)x

(3/2)x^2 - (3/2)x^3 - (1/2) x^2 + x^3 - (1/2) x^4 - x^3 + x^4 + (1/6) x^3 - (1/2) x^2 + (1/2) x - (1/6) - x + x^2 + (1/2) - (1/2)x

(3/2)x^2 - (4/3)x^3 + (1/2) x^4 + (1/3) - x

1

∫ ( (3/2)x^2 - (4/3)x^3 + (1/2) x^4 + (1/3) - x ) dx

0

. . . . . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. .1

(1/2)x^3 - (1/3)x^4 + (1/10) x^5 + (1/3)x - (1/2) x^2 ) ]

. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .0

(1/2) * ( 1^3 - 0^3 ) - (1/3) * ( 1^4 - 0^4 ) + (1/10) * ( 1^5 - 0^5 ) + (1/3) * (1 - 0) - (1/2) ( 1^2 - 0^2 )

(1/2) * ( 1 - 0 ) - (1/3) * ( 1 - 0 ) + (1/10) * ( 1 - 0 ) + (1/3) * 1 - (1/2) ( 1 - 0 )

(1/2) * 1 - (1/3) * 1 + (1/10) * 1 + (1/3) - (1/2) 1

(1/2) - (1/3) + (1/10) + (1/3) - (1/2)

1/10

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