Calc 4 ### Surface Integral CalculationTo evaluate the surface integral \[ \iint_S \sqrt{9 + 3z^2} \, dS \]for the surface $ S $ defined by the parametric equations:\[ \mathbf{r}(u, v) = \left[ 6 \cos(u) \sin(v), 6 \sin(u) \sin(v), 3 \cos(v) \right] \]we must consider the given bounds for the parameters $ u $ and $ v $:\[ 0 \leq u \leq \frac{\pi}{2}, \quad 0 \leq v \leq \frac{\pi}{2}. \]### Explanation of Parameters and Surface- $ \mathbf{r}(u, v) $ specifies the parameterization of the surface $ S $.- $ u $ and $ v $ are the parameters that vary within the given intervals.- $ 6 \cos(u) \sin(v) $ and $ 6 \sin(u) \sin(v) $ represent the x and y coordinates respectively.- $ 3 \cos(v) $ represents the z coordinate. The aim is to integrate the given function $ \sqrt{9 + 3z^2} $ over the surface area defined by the parametric equations.### Important ConsiderationsFor this problem, understanding the parametric representation of the surface and being able to compute the surface element $ dS $ will be crucial for carrying out the surface integral. The function under the integral involves the z-component of the parametric equations, which needs to be accounted for during the integration process.

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Description: Calc 4 ### Surface Integral CalculationTo evaluate the surface integral \[ \iint_S \sqrt{9 + 3z^2} \, dS \]for the surface $ S $ defined by the parametric equations:\[ \mathbf{r}(u, v) = \left[ 6 \cos(u) \sin(v), 6 \sin(u) \sin(v), 3 \cos(v) \right

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Answered: I. Evaluate '9 + 3z°dS for the surface…

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I. Evaluate '9 + 3z°dS for the surface S defined by S r(u, v) = [6 cos(u) sin(v), 6 sin(u) sin(v), 3 cos(v)] for 0

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Calc 4

### Surface Integral Calculation

To evaluate the surface integral

\[ \iint_S \sqrt{9 + 3z^2} \, dS \]

for the surface $ S $ defined by the parametric equations:

\[ \mathbf{r}(u, v) = \left[ 6 \cos(u) \sin(v), 6 \sin(u) \sin(v), 3 \cos(v) \right] \]

we must consider the given bounds for the parameters $ u $ and $ v $:

\[ 0 \leq u \leq \frac{\pi}{2}, \quad 0 \leq v \leq \frac{\pi}{2}. \]

### Explanation of Parameters and Surface

- $ \mathbf{r}(u, v) $ specifies the parameterization of the surface $ S $.
- $ u $ and $ v $ are the parameters that vary within the given intervals.
- $ 6 \cos(u) \sin(v) $ and $ 6 \sin(u) \sin(v) $ represent the x and y coordinates respectively.
- $ 3 \cos(v) $ represents the z coordinate.

The aim is to integrate the given function $ \sqrt{9 + 3z^2} $ over the surface area defined by the parametric equations.

### Important Considerations

For this problem, understanding the parametric representation of the surface and being able to compute the surface element $ dS $ will be crucial for carrying out the surface integral. The function under the integral involves the z-component of the parametric equations, which needs to be accounted for during the integration process.

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