Use the given transformation to evaluate the integral.
-$$\begin{array} { l }
x = 9 u , y = 10 v , z = 4 w ; \
\quad \int _ { R } \iint \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) d x d y d z
\end{array}$$
where $R$ is the interior of the ellipsoid $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 100 } + \frac { z ^ { 2 } } { 16 } = 1$
A) $144 \pi$
B) $216 \pi$
C) $240 \pi$
D) $288 \pi$

Question

Use the given transformation to evaluate the integral.
-$$\begin{array} { l } 
x = 9 u , y = 10 v , z = 4 w ; \
\quad \int _ { R } \iint \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) d x d y d z
\end{array}$$
where $R$ is the interior of the ellipsoid $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 100 } + \frac { z ^ { 2 } } { 16 } = 1$
A) $144 \pi$
B) $216 \pi$
C) $240 \pi$
D) $288 \pi$

Quizplus · Accepted Answer

The transformation simplifies the ellipsoid to a sphere $u^2 + v^2 + w^2 = 1$. The Jacobian of the transformation is $360$. The integral becomes $\int_0^1 \int_0^{2\pi} \int_0^{\pi} (81u^2 + 100v^2 + 16w^2) \cdot 360 \cdot u^2 \sin\theta \, du \, d\theta \, d\phi$, which evaluates to $288\pi$.

Use the Given Transformation to Evaluate the Integral Where RRR Is the Interior of the Ellipsoid

Use the Given Transformation to Evaluate the Integral Where $R$ Is the Interior of the Ellipsoid