Question

Surface Area The surface of the dome on a new museum is given by \(\mathbf{r}(u, v)=20 \sin u \cos v \mathbf{i}+20 \sin u \sin v \mathbf{j}+20 \cos u \mathbf{k}\) where \(0 \leq u \leq \pi / 3\) and \(0 \leq v \leq 2 \pi,\) and \(\mathbf{r}\) is in meters. Find the surface area of the dome.

Short answer

The surface area of the dome is \(4000 \pi m^2\).

Step-by-step solution

Compute the derivative of \(\mathbf{r}\) with respect to \(u\) and \(v\)

We know that \(\mathbf{r}(u, v) = 20 \sin(u) cos(v) \mathbf{i} + 20 \sin(u) \sin(v) \mathbf{j} + 20 \cos(u) \mathbf{k}\) . Let's compute \(\mathbf{r}_u\) and \(\mathbf{r}_v\) which are the partial derivative of \(\mathbf{r}\) with respect to \(u\) and \(v\) respectively.\[\mathbf{r}_u = 20 cos(u) cos(v) \mathbf{i} + 20 cos(u) \sin(v) \mathbf{j} - 20 sin(u) \mathbf{k}\]\[\mathbf{r}_v = -20 sin(u) sin(v) \mathbf{i} + 20 sin(u) cos(v) \mathbf{j}\]

Compute cross product of \(\mathbf{r}_u\) and \(\mathbf{r}_v\)

The cross product of \(\mathbf{r}_u\) and \(\mathbf{r}_v\) is given by: \(\mathbf{r}_u \times \mathbf{r}_v = (400 sin^2(u) - 400 sin^2(u), -400 sin(u) cos(u) sin(v), 400 sin(u) cos(u) cos(v)) = (0, -400 sin(u) cos(u) sin(v), 400 sin(u) cos(u) cos(v))\)

Compute the magnitude of the cross product

The magnitude of the cross product is given by \(\|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{(0)^2 + (-400 sin(u) cos(u) sin(v))^2 + (400 sin(u) cos(u) cos(v))^2} = 400 sin(u) cos(u) \).

Computing the surface area

The surface area \(A\) is given by the double integral of the magnitude of the cross product over the domain of \(u\) and \(v\). \[A = \int_0^{2 \pi} \int_0^{\frac{\pi}{3}} 400 \sin(u) \cos(u) dudv = 8000 \pi [0 - \frac{1}{2}]\]