# Find the region D for which the triple integral is a maximum and evaluate the integral on the...

## Question:

Find the region {eq}D {/eq} for which the triple integral is a maximum and evaluate the integral on the region:

{eq}\displaystyle \iiint_D (2-(x^2+y^2+2y+2z^2+4z) dV {/eq}

## Triple Integral:

A method of integration of the triple integral is using the iteration technique, i.e., integrating with respect to all three variables one by one within the specified limits of respective bounds. The triple integral is also known by the name volume Integral. Maximizing the volume integral is basically achieved by imposing certain conditions on the integrand and hence over the region of integration.

To find the region {eq}D {/eq} for which the triple integral {eq}\displaystyle \iiint_D (2-(x^2+y^2+2y+2z^2+4z) dV {/eq} is a maximum:

We need to keep the expression {eq}2-(x^2+y^2+2y+2z^2+4z)>0 {/eq}, in order to achieve the desired condition of greatest volume,

Hence,

{eq}\displaystyle x^2+(y+1)^2+2(z+1)^2<5 {/eq}

{eq}\Rightarrow D {/eq} must be the region inside the ellipsoid

{eq}\displaystyle \dfrac{x^2}5+\dfrac{(y+1)^2}5+\dfrac{(z+1)^2}{\frac 52}=1 {/eq}

Now in this region {eq}D {/eq} for which the triple integral is a maximum, the value of the integral on the region shall be:

{eq}\begin{align} I&=\displaystyle \int_{-\sqrt5}^{\sqrt5}\int_{-\sqrt5-1}^{\sqrt5-1}\int_{-\sqrt{5/2}-1}^{\sqrt{5/2}-1} (2-(x^2+y^2+2y+2z^2+4z) dx \,dy \, dz \\[2ex] &= \displaystyle \int_{-\sqrt5}^{\sqrt5}\int_{-\sqrt5-1}^{\sqrt5-1} \left(4\sqrt{\dfrac52}-2\sqrt{\dfrac52}\left(x^2+y^2+2y \right)+\frac43\left(-\sqrt{\frac 52}\right)^3+12\sqrt{\frac 52}-8\sqrt{\frac 52}\right)dx \, dy \\[2ex] &= \displaystyle \int_{-\sqrt5}^{\sqrt5}\left(2\sqrt5\left(\frac{14}3\sqrt{\frac52}-2\sqrt{\frac52}x^2\right)+\frac13\left(10\sqrt5+6\sqrt5\right)-4\sqrt5\right)dx \\[2ex] \therefore I&=\dfrac{140\sqrt{10}+40-50\sqrt2}3 \text{ cubic units} \end{align} {/eq} . 