Solve the problem.
-Find a parametrization for the ellipsoid $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 81 } + \frac { z ^ { 2 } } { 4 } = 1$. (Recall that the parametrization of an ellipse $\frac { x ^ { 2 } } { 36 } +$ $\frac { y ^ { 2 } } { 81 } = 1$ is $\left. x = 6 \cos \theta , y = 9 \sin \theta , 0 \leq \theta

Question

Solve the problem.
-Find a parametrization for the ellipsoid $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 81 } + \frac { z ^ { 2 } } { 4 } = 1$. (Recall that the parametrization of an ellipse $\frac { x ^ { 2 } } { 36 } +$ $\frac { y ^ { 2 } } { 81 } = 1$ is $\left. x = 6 \cos \theta , y = 9 \sin \theta , 0 \leq \theta < 2 \pi \right)$

Quizplus · Accepted Answer

The Answer of Solve the problem.
-Find a parametrization for the ellipsoid $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 81 } + \frac { z ^ { 2 } } { 4 } = 1$. (Recall that the parametrization of an ellipse $\frac { x ^ { 2 } } { 36 } +$ $\frac { y ^ { 2 } } { 81 } = 1$ is $\left. x = 6 \cos \theta , y = 9 \sin \theta , 0 \leq \theta < 2 \pi \right)$

Solve the Problem x236+y281+z24=1\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 81 } + \frac { z ^ { 2 } } { 4 } = 136x2​+81y2​+4z2​=1

Solve the Problem $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 81 } + \frac { z ^ { 2 } } { 4 } = 1$