Find the area of the surface: the helicoid (or spiral ramp) with vector equation: $\mathbf{r}(u, v) = u \cos(v) \mathbf{i} + u \sin(v) \mathbf{j} + v \mathbf{k},$ where $0 \leq u \leq 1$ and $0 \leq v \leq 5\pi$.

Neetesh Kumar | November 16, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 16.6 Question Number 23
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Step-by-step solution:

To find the surface area of a parametric surface, we use the formula: $A = \iint_R \|\mathbf{r}_u \times \mathbf{r}_v\| \, dA$

where:

• $\mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u}$ (partial derivative with respect to $u$),
• $\mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v}$ (partial derivative with respect to $v$),
• $R$ is the region in the $uv$-plane defined by $0 \leq u \leq 1$, $0 \leq v \leq 5\pi$.

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Step 1: Compute $\mathbf{r}_u$ and $\mathbf{r}_v$

The parametric equation is: $\mathbf{r}(u, v) = \langle u \cos(v), u \sin(v), v \rangle$

1. Compute $\mathbf{r}_u$:

   • $\frac{\partial}{\partial u}(u \cos(v)) = \cos(v)$,
   • $\frac{\partial}{\partial u}(u \sin(v)) = \sin(v)$,
   • $\frac{\partial}{\partial u}(v) = 0$.

   Thus: $\mathbf{r}_u = \langle \cos(v), \sin(v), 0 \rangle$

2. Compute $\mathbf{r}_v$:

   • $\frac{\partial}{\partial v}(u \cos(v)) = -u \sin(v)$,
   • $\frac{\partial}{\partial v}(u \sin(v)) = u \cos(v)$,
   • $\frac{\partial}{\partial v}(v) = 1$.

   Thus: $\mathbf{r}_v = \langle -u \sin(v), u \cos(v), 1 \rangle$

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Step 2: Compute the Cross Product $\mathbf{r}_u \times \mathbf{r}_v$

The cross product $\mathbf{r}_u \times \mathbf{r}_v$ is: $\mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos(v) & \sin(v) & 0 \\ -u \sin(v) & u \cos(v) & 1 \end{vmatrix}$

Expanding the determinant:

1. For $\mathbf{i}$: $\mathbf{i} \begin{vmatrix} \sin(v) & 0 \\ u \cos(v) & 1 \end{vmatrix} = \mathbf{i} \left[\sin(v) - 0\right] = \mathbf{i} \sin(v)$

2. For $\mathbf{j}$: $-\mathbf{j} \begin{vmatrix} \cos(v) & 0 \\ -u \sin(v) & 1 \end{vmatrix} = -\mathbf{j} \left[\cos(v)(1) - (0)(-u \sin(v))\right] = -\mathbf{j} \cos(v)$

3. For $\mathbf{k}$: $\mathbf{k} \begin{vmatrix} \cos(v) & \sin(v) \\ -u \sin(v) & u \cos(v) \end{vmatrix} = \mathbf{k} \left[u \cos^2(v) + u \sin^2(v)\right] = \mathbf{k} u$

Thus: $\mathbf{r}_u \times \mathbf{r}_v = \langle \sin(v), -\cos(v), u \rangle$

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Step 3: Compute the Magnitude of $\mathbf{r}_u \times \mathbf{r}_v$

The magnitude is: $\|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{\sin^2(v) + (-\cos(v))^2 + u^2}$

Simplify: $\|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{\sin^2(v) + \cos^2(v) + u^2} = \sqrt{1 + u^2}$

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Step 4: Set Up the Double Integral

The surface area is: $A = \iint_R \sqrt{1 + u^2} \, dA$

Substitute the bounds for $u$ and $v$: $A = \int_0^{5\pi} \int_0^1 \sqrt{1 + u^2} \, du \, dv$

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Step 5: Evaluate the Inner Integral (with respect to $u$)

Let: $I = \int_0^1 \sqrt{1 + u^2} \, du$

Use the substitution $u = \sinh(t)$, so $du = \cosh(t) \, dt$ and $1 + u^2 = \cosh^2(t)$.

When $u = 0$, $t = 0$, and when $u = 1$, $t = \sinh^{-1}(1)$.

The integral becomes: $I = \int_0^{\sinh^{-1}(1)} \cosh^2(t) \, dt$

Use the identity $\cosh^2(t) = \frac{1}{2}(\cosh(2t) + 1)$: $I = \frac{1}{2} \int_0^{\sinh^{-1}(1)} (\cosh(2t) + 1) \, dt$

Separate the terms: $I = \frac{1}{2} \int_0^{\sinh^{-1}(1)} \cosh(2t) \, dt + \frac{1}{2} \int_0^{\sinh^{-1}(1)} 1 \, dt$

For $\int \cosh(2t) \, dt$, the integral is $\frac{1}{2} \sinh(2t)$. For $\int 1 \, dt$, the result is $t$.

Evaluate: $I = \frac{1}{2} \left[\frac{1}{2} \sinh(2t) \right]_0^{\sinh^{-1}(1)} + \frac{1}{2} \left[t\right]_0^{\sinh^{-1}(1)}$ $I = \frac{1}{4} \sinh(2 \sinh^{-1}(1)) + \frac{1}{2} \sinh^{-1}(1)$

Simplify using $\sinh(2x) = 2\sinh(x)\cosh(x)$ and $\sinh^{-1}(1) = \ln(1 + \sqrt{2})$.

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Step 6: Evaluate the Outer Integral (with respect to $v$)

After evaluating $I$, multiply by the $v$-integral: $\int_0^{5\pi} dv = 5\pi$

Thus: $A = 5\pi \cdot I$

Final Answer:

The surface area is: $A = 5\pi \left[\frac{1}{4} \sinh(2 \sinh^{-1}(1)) + \frac{1}{2} \sinh^{-1}(1)\right]$

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