Question

Math

Posted 5 months ago

```
Suppose we have a surface $S$ defined by the transformation $T$ for $\frac{-\pi}{2}<u<\frac{\pi}{2}$ and $-1<v<1$.
$T(u, v)=(\cos (u), \sin (u), v)$
What is the surface area of $S$ ?
Choose 1 answer:
(A) $\pi$
(B) $2 \pi$
(C) $3 \pi$
(D) $4 \pi$
```

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Answer from Sia

Posted 5 months ago

Solution by Steps

step 2

First, we find the partial derivatives of $T(u, v)$: $T_u(u, v) = (-\sin(u), \cos(u), 0)$ and $T_v(u, v) = (0, 0, 1)$

step 3

Next, we compute the cross product of $T_u(u, v)$ and $T_v(u, v)$: $T_u \times T_v = (\cos(u), \sin(u), 0) \times (0, 0, 1) = (\sin(u), -\cos(u), 0)$

step 4

The magnitude of the cross product is $|T_u \times T_v| = \sqrt{\sin^2(u) + (-\cos(u))^2} = \sqrt{1}$, since $\sin^2(u) + \cos^2(u) = 1$

step 5

The surface area of $S$ is the double integral of the magnitude of the cross product over the given range: $\int_{-1}^{1} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} |T_u \times T_v| \, du \, dv$

step 6

Since $|T_u \times T_v| = 1$, the integral simplifies to $\int_{-1}^{1} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, du \, dv$

step 7

Evaluating the integral, we get $\int_{-1}^{1} \left[ u \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} dv = \int_{-1}^{1} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) dv$

step 8

This simplifies to $\int_{-1}^{1} \pi \, dv = \pi \left[ v \right]_{-1}^{1} = \pi (1 - (-1)) = 2\pi$

step 9

Therefore, the surface area of $S$ is $2\pi$

B

Key Concept

Surface Area of a Parametric Surface

Explanation

The surface area of a parametric surface can be found by integrating the magnitude of the cross product of the partial derivatives of the transformation over the given range.

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