Suppose we have a surface $S$ defined by the transformation $T$ for $\frac{-\pi}{2}

Q: 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) $4 \pi$

∻Solution by Steps∻ ‖ step 1 ⋮ To find the surface area of $S$, we need to compute the integral of the magnitude of the cross product of the partial derivatives of $T(u, v)$ with respect to $u$ and $v$ over the given range. ‖ ‖ 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.⚹◊What is the formula to calculate the surface area of a surface defined by a given transformation in terms of u and v?⍭ Generate me a similar question◊

Question

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) $4 \pi$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the surface area of $S$, we need to compute the integral of the magnitude of the cross product of the partial derivatives of $T(u, v)$ with respect to $u$ and $v$ over the given range. ‖ ‖ 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 $T_v(u, v)$: $T_u imes T_v = (\cos(u), \sin(u), 0) imes (0, 0, 1) = (\sin(u), -\cos(u), 0)$. ‖ ‖ step 4 ⋮ The magnitude of the cross product is $|T_u imes T_v| = \sqrt{\sin^2(u) + (-\cos(u))^2} = \sqrt{1}$, 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: $\int_{-1}^{1} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} |T_u imes T_v| \, du \, dv$. ‖ ‖ step 6 ⋮ Since $|T_u imes T_v| = 1$, the integral simplifies to $\int_{-1}^{1} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, du \, dv$. ‖ ‖ step 7 ⋮ Evaluating the integral, we get $\int_{-1}^{1} \left[ u ight]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} dv = \int_{-1}^{1} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2} ight) ight) dv$. ‖ ‖ step 8 ⋮ This simplifies to $\int_{-1}^{1} \pi \, dv = \pi \left[ v ight]_{-1}^{1} = \pi (1 - (-1)) = 2\pi$. ‖ ‖ 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.⚹◊What is the formula to calculate the surface area of a surface defined by a given transformation in terms of u and v?⍭ Generate me a similar question◊

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