# Evaluate Triple Integral from R of the square root of [1 - ((x^2)/36 + (y^2)/81 + (z^2)/16)]...

## Question:

Evaluate

• {eq}\displaystyle{\iiint_R \, \sqrt{1 \, - \, \left ( \frac{x^{2}}{36} \, + \, \frac{y^{2}}{81} \, + \, \frac{z^{2}}{16} \right )} \, dx \, dy \, dz} {/eq},

where {eq}\, R \, {/eq} is the interior of the ellipsoid {eq}\, \displaystyle{\frac{x^{2}}{36} \, + \, \frac{y^{2}}{81} \, + \, \frac{z^{2}}{16} \, = \, 1} {/eq}. Hint: Let {eq}\, x \, = \, 6u, \quad y \, = \, 9v, \quad z \, = \, 4w, {/eq} and then convert to spherical coordinates.

## Change of Variables:

We use the Jacobian to change variables in a triple integral which is given by:

{eq}J =\dfrac{\partial (x,y,z)}{\partial (u,v,w)}=\begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w}\\ \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w} \end{vmatrix}\\ {/eq}

The formula that we used to compute the triple integral is:

{eq}\iiint_{R}f(x,y,z)dxdy dz=\iiint_{D}f(g(u,v,w), h(u,v,w), p(u,v,w))\left | J \right |dudvdw {/eq}

We have to evaluate the integral {eq}\int\iint_R \sqrt{1-\left (\dfrac{x^2}{36}+\dfrac{y^2}{81}+\dfrac{z^2}{16} \right )}\;dx\; dy\; dz {/eq}, where R is the intersection of the ellipsoid {eq}\dfrac{x^2}{36}+\dfrac{y^2}{81}+\dfrac{z^2}{16} =1 {/eq}.

The given transformation is {eq}x=6u, y = 9v, z=4w {/eq}

To evaluate the integral, we use the following formula:

{eq}\iiint_{R}f(x,y,z)dxdy dz=\iiint_{D}f(g(u,v,w), h(u,v,w), p(u,v,w))\left | J \right |dudvdw {/eq}

The original domain of the integral is bounded by the ellipse {eq}\dfrac{x^2}{36}+\dfrac{y^2}{81}+\dfrac{z^2}{16} =1 {/eq}

After the transformation {eq}x=6u, y = 9v, z=4w {/eq}, the domain changes to the region bounded by:

{eq}\begin{align} \dfrac{(6u)^2}{36}+\dfrac{(9v)^2}{81}+\dfrac{(4w)^2}{16} =1\\ u^2+v^2+w^2 &=1 \end{align} {/eq}

So the new domain becomes the unit sphere {eq}u^2+v^2+w^2=1 {/eq}

Now, we calculate the Jacobian.

{eq}\begin{align} J &=\dfrac{\partial (x,y,z)}{\partial (u,v,w)}=\begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w}\\ \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w} \end{vmatrix}\\ &=\begin{vmatrix} 6 & 0 & 0\\ 0 & 9 & 0\\ 0 & 0 & 4 \end{vmatrix}\\ &=6(36-0)\\ &=216 \end{align} {/eq}

Therefore:

{eq}\begin{align} \int\iint_R \sqrt{1-\left (\frac{x^2}{36}+\frac{y^2}{81}+\frac{z^2}{16} \right )}\;dx\; dy\; dz &=\int\iint_R \sqrt{1-\left (u^2+v^2+w^2 \right )}(216)dudvdw\\ &=216\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}\sqrt{1-r^2}(r^2\sin (\phi))drd\phi d\theta\\ &=216\int_{0}^{2\pi}\int_{0}^{\pi}\sin (\phi)\left (\int_{0}^{1}r^2\sqrt{1-r^2}dr \right )d\phi d\theta\\ \\ & \begin{Bmatrix} Put \: r=\sin (u)\Rightarrow dr=\cos u \:du\\ \int_{0}^{1} r^2\sqrt{1-r^2}dr=\int_{0}^{\frac{\pi}{2}} \sin^2(u)\cos^2(u)du=\dfrac{1}{4}\int_{0}^{\frac{\pi}{2}} \sin^2(2u)du=\dfrac{1}{4}\int_{0}^{\frac{\pi}{2}}\left ( \dfrac{1-\cos (4u)}{2} \right )du=\dfrac{1}{8}(u-\dfrac{\sin (4u)}{4})_{0}^{\dfrac{\pi}{2}}=\dfrac{1}{8}\left ( \dfrac{\pi}{2}-\dfrac{\sin(2\pi)}{4} \right )=\dfrac{1}{8}\left ( \dfrac{\pi}{2} \right ) \\ \int r^2\sqrt{1-r^2}dr=\dfrac{\pi}{16} \end{Bmatrix}\\ \\ &=216\int_{0}^{2\pi}\int_{0}^{\pi}\sin (\phi)\left ( \frac{\pi}{16} \right )d\phi d\theta\\ &=\dfrac{216}{16}\pi\int_{0}^{2\pi}\left (\int_{0}^{\pi}\sin (\phi d\phi) \right )d\theta\\ &=\dfrac{216}{16}\pi\int_{0}^{2\pi}\left ( -\cos (\phi) \right )_{0}^{\pi}d\theta\\ &=\dfrac{216}{16}\pi\int_{0}^{2\pi}\left ( -\cos (\pi)-\cos (0) \right )d\theta\\ &=\dfrac{216}{16}\pi\int_{0}^{2\pi}\left ( 2 \right )d\theta\\ &=\dfrac{216}{8}\pi\int_{0}^{2\pi}d\theta\\ &=\dfrac{216}{8}\pi(\theta)_{0}^{2\pi}\\ &=\dfrac{216}{8}\pi(2\pi-0)\\ &=54\pi^2\\ \end{align} {/eq}

So, the correct option is {eq}\color{blue}{D)} {/eq}