Question

In Exercises 13-16, evaluate \(\int_{S} \int f(x, y) d S\). $$ \begin{array}{l} f(x, y)=x y ; S: \mathbf{r}(u, v)=2 \cos u \mathbf{i}+2 \sin u \mathbf{j}+v \mathbf{k} \\ 0 \leq u \leq \frac{\pi}{2}, \quad 0 \leq v \leq 2 \end{array} $$

Short answer

The value of the surface integral \(\int_{S} \int f(x, y) \, dS\) is \(16/3\).

Step-by-step solution

Express the function f(x, y) in terms of u and v

The vector function \(\mathbf{r}(u, v)\) is \(2 \cos u \mathbf{i} + 2 \sin u \mathbf{j} + v \mathbf{k}\). Here, \(x = 2\cos u\) and \(y = 2\sin u\). So the function \(f(x, y)\) in terms of \(u\) and \(v\) is \(f(u, v) = x*y = (2\cos u * 2\sin u)\).

Compute the cross product of the partial derivatives of \(\mathbf{r}(u, v)\)

Firstly, find the partial derivatives \(\mathbf{r_u}\) and \(\mathbf{r_v}\). \(\mathbf{r_u} = -2\sin u \mathbf{i} + 2\cos u \mathbf{j} + 0 * \mathbf{k}\) and \(\mathbf{r_v} = 0 * \mathbf{i} + 0 * \mathbf{j} + 1 * \mathbf{k}\). Now, compute the cross product: \(\mathbf{N} = \mathbf{r_u} \times \mathbf{r_v} = (0 - 0)\mathbf{i} - (2\sin u - 0)\mathbf{j} + (0 - 0)\mathbf{k} = -2\sin u * \mathbf{j}\).

Determine the magnitude of the cross product

The magnitude of a vector \(\mathbf{A} = A_x* \mathbf{i} + A_y* \mathbf{j} + A_z* \mathbf{k}\) is \(\sqrt{A_x^2 + A_y^2 + A_z^2}\). So the magnitude of the vector N is \(\|\mathbf{N}\| = \sqrt{(0)^2 + (-2\sin u)^2 + (0)^2} = |2\sin u|\).

Evaluate the double integral

The surface integral is found by evaluating the double integral of \(f(u, v) * \|\mathbf{N}\|\) over the given limits for \(u\) and \(v\). \(\int_{S} \int f \, dS = \int_0^2 \int_0^{\pi/2} (2\cos u * 2\sin u) * |2\sin u| \, du \, dv\). Since \(\sin u\) is positive for \(0 \leq u \leq \pi/2\), the integral simplifies to: \(\int_0^2 dv \int_0^{\pi/2} 8\sin^2(u)\cos(u) \, du = 8 \int_0^2 dv \int_0^{\pi/2} \sin^2(u)\cos(u) \, du = 8v\Big|_0^2 * [\sin^3(u)/3]\Big|_0^{\pi/2}\). Thus, the evaluated integral is \((8*2)*[1/3 - 0] = 16/3\) .