Evaluate the triple integral.


E
y dV, where E = {(x, y, z) | 0 ≤ x ≤ 6, 0 ≤ y ≤ x, x − y ≤ z ≤ x + y}

The definition of the set E gives you a natural choice for the limits in the integral:

[tex]\displaystyle \iiint_E y \, dV = \int_0^6 \int_0^x \int_{x-y}^{x+y} y \, dz \, dy \, dx[/tex]

Computing the integral, we get

[tex]\displaystyle \iiint_E y \, dV = \int_0^6 \int_0^x y ((x+y)-(x-y)) \, dy \, dx = 2 \int_0^6 \int_0^x y^2 \, dy \, dx[/tex]

[tex]\displaystyle \iiint_E y \, dV = 2 \int_0^6 \frac13 (x^3 - 0^3) \, dx = \frac23 \int_0^6 x^3 \, dx[/tex]

[tex]\displaystyle \iiint_E y \, dV = \frac23 \cdot \frac14 (6^4 - 0^4) = \boxed{216}[/tex]

The evaluation of the triple integral ∭E y dV, where E = {(x, y, z) | 0 ≤ x ≤ 3, 0 ≤ y ≤ x, x − y ≤ z ≤ x + y} is 216

Triple integral integrates the integrand over three dimensions.

Given that:

E = {(x, y, z) | 0 ≤ x ≤ 3, 0 ≤ y ≤ x, x − y ≤ z ≤ x + y}

Evaluating the integral:

[tex]I = \int \int \int _E y \: dV[/tex]

[tex]I = \mathlarger{\int_0^6 (\int_0^x (\int_{x-y}^{x+y} y\: dz)dy)dx}\\\\I = \int_0^6 (\int_0^x [yz]_{z=(x-y)}^{z=(x+y)}dy)dx\\\\I = \int_0^6 (\int_0^x y( (x+y) - (x-y)) \: dy)dx\\\\ I = \int_0^6 (\int_0^x y(2y) \: dy)dx\\\\I =2 \int_0^6 \left [\dfrac{y^3}{3}\right ]_{y=0}^{y=x} dx\\\\I = 2 \int_0^6 (x^3/3) dx\\\\I = \dfrac{2}{3} \left[\dfrac{x^4}{4} \right]_{x=0}^{x=6} = \dfrac{2}{3} \times \dfrac{6^4}{4} =216[/tex]

Thus, the evaluation of the triple integral ∭E y dV, where E = {(x, y, z) | 0 ≤ x ≤ 3, 0 ≤ y ≤ x, x − y ≤ z ≤ x + y} is 216

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