Question

(1 point) the vector equation r(u,v)=ucosvi usinvj vk, 0≤v≤6π, 0≤u≤1, describes a helicoid (spiral ramp). what is the surface area?

Answer 1

The surface area of the helicoid described by the vector equation
`\( \mathbf{r(u, v)} = u\cos(v) \mathbf{i} + u\sin(v) \mathbf{j} + v\mathbf{k} \)`, where
`\( 0 \leq v \leq 6\pi \) and \( 0 \leq u \leq 1 \),` is
`\( 18\pi \)`square units.

Step-by-step explanation:

The surface area of a parametrically defined surface \( S \) can be calculated using the surface area integral:

`\[ A = \iint_S \| \mathbf{r}_u * \mathbf{r}_v \| \, du \, dv \]`

where
`\( \mathbf{r}(u, v) \)` is the vector-valued function that describes the surface. In this case,
`\( \mathbf{r}(u, v) = u\cos(v) \mathbf{i} + u\sin(v) \mathbf{j} + v\mathbf{k} \)`.

To compute
`\( \mathbf{r}_u \)` and
`\( \mathbf{r}_v \),` we differentiate
`\( \mathbf{r}(u, v) \)` with respect to u and v , respectively. After obtaining these partial derivatives, we find their cross product and take its magnitude.

`\[ \mathbf{r}_u = \cos(v) \mathbf{i} + \sin(v) \mathbf{j} \]`

`\[ \mathbf{r}_v = -u\sin(v) \mathbf{i} + u\cos(v) \mathbf{j} + \mathbf{k} \]`

`\[ \mathbf{r}_u * \mathbf{r}_v = u\cos(v) \mathbf{i} + u\sin(v) \mathbf{j} + u \mathbf{k} \]`

The magnitude of this cross product is
`\( √(2)u \)` . The surface area integral then simplifies to
`\( A = \int_0^1 \int_0^(6\pi) √(2)u \, dv \, du \)`, which evaluates to \( 18\pi \) square units.
`\( 18\pi \)`

Answer 2

The surface area of the helicoid (spiral ramp) can be calculated using the vector equation and the formula for surface area. By calculating the partial derivatives and substituting them into the formula, we can find the surface area to be 6π.

Step-by-step explanation:

The surface area of a helicoid (spiral ramp) can be calculated using the formula:

A = ∫∫|r′(u,v) × r′′(u,v)| dudv

where r′(u, v) and r′′(u, v) are the partial derivatives of the vector equation r(u, v) with respect to u and v, respectively. In this case, r(u, v) = ucos(v)i + usin(v)j + vk.

By calculating the partial derivatives, we obtain:

r′(u, v) = cos(v)i + sin(v)j + 0k

r′′(u, v) = 0i + 0j + 1k

Substituting these values into the formula, we get:

A = ∫∫|(cos(v)i + sin(v)j + 0k) × (0i + 0j + 1k)| dudv = ∫∫|j| dudv

Since |j| = 1, the integral becomes:

A = ∫∫ dudv = ∫v=0..6π du = 6πu

Finally, substituting the limits of u (0 to 1), we get:

A = 6π