Problem 37 - Calculus Early Transcendentals

Question

Evaluate the integral \(\iint_{\sigma} f(x, y, z) d S\) over the surface \(\sigma\) represented by the vector-valued function \(\mathbf{r}(u, v) .\) $$ \begin{array}{l}{f(x, y, z)=\frac{1}{\sqrt{1+4 x^{2}+4 y^{2}}}} \\\ {\mathbf{r}(u, v)=u \cos v \mathbf{i}+u \sin v \mathbf{j}+u^{2} \mathbf{k}} \\\ {(0 \leq u \leq \sin v, 0 \leq v \leq \pi)}\end{array} $$

Step-by-Step Solution

Answer

The value of the surface integral is \(\frac{\pi}{4}\).

Step 1: Determine the Parametric Surface

The surface \(\sigma\) is represented by the parametric equations \(x = u \cos v\), \(y = u \sin v\), and \(z = u^2\), with parameters \(0 \leq u \leq \sin v\) and \(0 \leq v \leq \pi\).

Step 2: Compute the Partial Derivatives

To find the differential surface area element \(dS\), compute the partial derivatives of \(\mathbf{r}(u,v)\) with respect to \(u\) and \(v\):\[\frac{\partial \mathbf{r}}{\partial u} = \cos v \mathbf{i} + \sin v \mathbf{j} + 2u \mathbf{k}\]\[\frac{\partial \mathbf{r}}{\partial v} = -u \sin v \mathbf{i} + u \cos v \mathbf{j}\]

Step 3: Calculate the Cross Product

Compute the cross product \(\mathbf{r}_u \times \mathbf{r}_v\):\[\mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \cos v & \sin v & 2u \ -u \sin v & u \cos v & 0 \end{vmatrix}\]Simplifying the determinant, we get:\[\mathbf{r}_u \times \mathbf{r}_v = (-2u^2 \cos v) \mathbf{i} + (-2u^2 \sin v) \mathbf{j} + u \mathbf{k}\]

Step 4: Find the Magnitude of the Cross Product

Calculate the magnitude of the cross product vector:\[\|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{(-2u^2 \cos v)^2 + (-2u^2 \sin v)^2 + u^2}\]This simplifies to:\[\|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{4u^4 + u^2} = u\sqrt{4u^2 + 1}\]

Step 5: Set Up the Surface Integral

The integral becomes:\[\iint_\sigma f(x, y, z) \, dS = \int_0^\pi \int_0^{\sin v} \frac{1}{\sqrt{1 + 4(u \cos v)^2 + 4(u \sin v)^2}}\cdot u\sqrt{4u^2+1} \, du \, dv\]This simplifies further to:\[= \int_0^\pi \int_0^{\sin v} \frac{u\sqrt{4u^2+1}}{\sqrt{1 + 4u^2}} \, du \, dv\]

Step 6: Simplify and Integrate

Recognizing the simplification \(\sqrt{1 + 4u^2}\), the expression simplifies to \(u\):\[\iint_\sigma \frac{u\sqrt{4u^2+1}}{\sqrt{1 + 4u^2}} \, du \, dv = \int_0^\pi \int_0^{\sin v} u \, du \, dv\]This is now a straightforward double integral:\[\int_0^\pi \left[ \frac{u^2}{2} \right]_0^{\sin v} \, dv = \int_0^\pi \frac{\sin^2 v}{2} \, dv\]

Step 7: Evaluate the Integral

Finally, the integral of \(\frac{\sin^2 v}{2}\) is evaluated. Recall that \(\sin^2 v = \frac{1 - \cos(2v)}{2}\):\[= \int_0^\pi \frac{1 - \cos(2v)}{4} \, dv\]This results in:\[= \left[ \frac{v}{4} - \frac{\sin(2v)}{8} \right]_0^\pi = \left[ \frac{\pi}{4} - 0 \right]\]Thus, the value of the surface integral is \(\frac{\pi}{4}\).

Key Concepts

Parametric Surfaces

Parametric surfaces are a way to represent a surface in three-dimensional space using two parameters, typically denoted as \(u\) and \(v\). Instead of relying on traditional Cartesian coordinates \((x, y, z)\), parametric surfaces describe each point on the surface based on parameter values. This method is powerful because it allows for the easy representation and manipulation of complex surfaces.

In our exercise, we have the vector-valued function \(\mathbf{r}(u, v) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^2 \mathbf{k}\). This function maps any point \((u, v)\) to a point \((x, y, z)\) in 3D space:

• \(x = u \cos v\)
• \(y = u \sin v\)
• \(z = u^2\)

This specific surface is described over the range \(0 \leq u \leq \sin v\) and \(0 \leq v \leq \pi\).

By adjusting the parameters, you can trace out every point on the surface, which is particularly useful in calculating integrals over those surfaces.

Cross Product

The cross product is a mathematical operation used to find a vector perpendicular to two given vectors in three-dimensional space. It's particularly useful in determining the normal vector to a surface, which is essential for calculating surface integrals.

In this context, the cross product \(\mathbf{r}_u \times \mathbf{r}_v\) is calculated from the partial derivatives \(\frac{\partial \mathbf{r}}{\partial u}\) and \(\frac{\partial \mathbf{r}}{\partial v}\):

• \(\frac{\partial \mathbf{r}}{\partial u} = \cos v \mathbf{i} + \sin v \mathbf{j} + 2u \mathbf{k}\)
• \(\frac{\partial \mathbf{r}}{\partial v} = -u \sin v \mathbf{i} + u \cos v \mathbf{j}\)

Using the determinant formula for vectors, we find:
\[\mathbf{r}_u \times \mathbf{r}_v = (-2u^2 \cos v) \mathbf{i} + (-2u^2 \sin v) \mathbf{j} + u \mathbf{k}\]

The cross product results in a vector that is oriented perpendicular to the surface at each point, crucial for further calculations.

Double Integral

A double integral is a way to integrate over a two-dimensional region. In the context of surfaces, the double integral calculates the accumulation of a quantity over a surface area.

In our exercise, the double integral expression is:
\[\iint_{\sigma} f(x, y, z) \, dS\]
This integral expression uses the function \(f(x, y, z)\) and the differential element \(dS\), which represents an infinitesimal piece of the surface. The value of the surface integral gives a sense of the function's "content" over the surface.

The setup for the double integral is:
\[\int_0^\pi \int_0^{\sin v} \frac{u\sqrt{4u^2+1}}{\sqrt{1 + 4u^2}} \, du \, dv\]

This process allows us to determine how much of the function is accumulated over the specific range of the parameters.

Partial Derivatives

Partial derivatives are derivatives of functions with multiple variables where one variable is differentiated at a time. They are used extensively in calculus, particularly when dealing with parametrically defined surfaces.

For our surface defined by \(\mathbf{r}(u, v)\), we find the partial derivatives with respect to \(u\) and \(v\) to help determine the area element \(dS\):

• \(\frac{\partial \mathbf{r}}{\partial u} = \cos v \mathbf{i} + \sin v \mathbf{j} + 2u \mathbf{k}\)
• \(\frac{\partial \mathbf{r}}{\partial v} = -u \sin v \mathbf{i} + u \cos v \mathbf{j}\)

The partial derivatives represent tangent vectors to the surface for constant \(u\) and \(v\), respectively. They help form the base vectors from which the normal vector \(\mathbf{r}_u \times \mathbf{r}_v\) is computed, which then assists in calculating the surface element \(dS\).

Understanding partial derivatives is fundamental when calculating surface integrals over parametric surfaces, highlighting the strong connection between multivariable calculus and surface geometry.