Question

How to evaluate the triple integral with step by step solution please?

Answer 1

`int_1^2 \ int_0^x \ int _0^(1-y) \ x^3y^2z \ dz \ dy \ dx = 7387/10080`

Explanation:

We seek:

`I = int_1^2 \ int_0^x \ int _0^(1-y) \ x^3y^2z \ dz \ dy \ dx`

We evaluate multiple integrals from the inner to the outer and integrate wrt to the appropriate variable whilst treating all others as constant, thus for the first integration, showing all steps we have:

`I = int_1^2 \ int_0^x \ {int _0^(1-y) \ x^3y^2z \ dz } \ dy \ dx`

`\ \ = int_1^2 \ int_0^x \ x^3y^2 \ {int _0^(1-y) \ z \ dz } \ dy \ dx`

`\ \ = int_1^2 \ int_0^x \ x^3y^2 \ [ \ z^2/2 ] _(z=0)^(z=1-y) \ dy \ dx`

`\ \ = int_1^2 \ int_0^x \ (x^3y^2)/2 \ ( (1-y)^2 - 0^2) \ dy \ dx`

`\ \ = int_1^2 \ int_0^x \ (x^3y^2)/2 \ (1-y)^2 \ dy \ dx`

`\ \ = int_1^2 \ x^3/2 \ int_0^x \ y^2(1-y)^2 \ dy \ dx`

`\ \ = int_1^2 \ x^3/2 \ int_0^x \ y^2-2y^3+y^4 \ dy \ dx`

`\ \ = int_1^2 \ x^3/2 \ [ \ y^3/3-y^4/2+y^5/5 ]_(y=0)^(y=x) \ dy \ dx`

`\ \ = int_1^2 \ x^3/2 \ { (x^3/3-x^4/2+x^5/5) - (0) } \ dx`

`\ \ = int_1^2 \ x^3/2(x^3/3-x^4/2+x^5/5) \ dx`

`\ \ = 1/2 \ int_1^2 \ x^6/3-x^7/2+x^8/5 \ dx`

`\ \ = 1/2 \ [ \ x^7/21-x^8/16+x^9/45 \ ]_1^2`

`\ \ = 1/2 \ { ( 2^7/21-2^8/16+2^9/45) - (1/21-1/16+1/45) }`

`\ \ = 1/2 \ { ( 128/21-256/16+512/45) - (1/21-1/16+1/45) }`

`\ \ = 1/2 \ (464/315 - 37/5040)`

`\ \ = 1/2 \ 7387/5040`

`\ \ = 7387/10080`