Problem 8 - Advanced Engineering Mathematics

Question

Find the surface area of that portion of the graph of \(z=x^{2}-y^{2}\) that is in the first octant within the cylinder \(x^{2}+y^{2}=4\)

Step-by-Step Solution

Answer

The surface area is \(\frac{\pi}{24}(17^{3/2} - 1)\).

Step 1: Understand the problem setup

We need to find the surface area of the portion of the surface given by \( z = x^2 - y^2 \) that lies within the first octant and inside the cylinder \( x^2 + y^2 = 4 \). Remember, the first octant implies that \( x, y, z \geq 0 \).

Step 2: Set up the surface area integral

The surface area of a function \( z = f(x,y) \) can be found using the formula: \[ A = \iint_{D} \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dA \] where \( D \) is the domain over which the surface exists.

Step 3: Compute partial derivatives

Calculate the partial derivatives of \( z = x^2 - y^2 \). We have: \[ \frac{\partial z}{\partial x} = 2x \] and \[ \frac{\partial z}{\partial y} = -2y \].

Step 4: Substitute into the surface area formula

Substitute the derivatives into the square root expression: \[ \sqrt{1 + (2x)^2 + (-2y)^2} = \sqrt{1 + 4x^2 + 4y^2} \] simplifying to \[ \sqrt{1 + 4(x^2 + y^2)} \].

Step 5: Set bounds for the integral

Since we are dealing with the first octant inside the cylinder, the bounds for \( x \) and \( y \) are \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq \sqrt{4-x^2} \) due to \( x^2 + y^2 = 4 \).

Step 6: Convert to polar coordinates

Convert the problem to polar coordinates where \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). Therefore, \( x^2 + y^2 = r^2 \), and \( 0 \leq \theta \leq \frac{\pi}{2} \) for the first octant. \( 0 \leq r \leq 2 \).

Step 7: Evaluate the double integral

The surface area in polar coordinates becomes: \[ A = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \sqrt{1 + 4r^2} \cdot r \, dr \, d\theta \]. Evaluate the integral with respect to \( r \) first, then \( \theta \).

Step 8: Solve the integral with respect to \( r \)

Calculate \( \int_{0}^{2} r\sqrt{1 + 4r^2} \, dr \). Let \( u = 1 + 4r^2 \), then \( du = 8r \, dr \), hence \( r \, dr = \frac{1}{8} \, du \). Substitute and integrate: \[ \int \frac{1}{8}u^{1/2} \, du = \frac{1}{12}(1 + 4r^2)^{3/2} \bigg|_{0}^{2} \].

Step 9: Simplify and evaluate the integral for \( r \)

Evaluate the integral: \[ \frac{1}{12}\left[(1 + 4(2)^2)^{3/2} - (1 + 4(0)^2)^{3/2}\right] = \frac{1}{12}(17^{3/2} - 1) \].

Step 10: Evaluate the integral with respect to \( \theta \)

Now integrate with respect to \( \theta \): \[ \int_{0}^{\frac{\pi}{2}} \frac{1}{12}(17^{3/2} - 1) \, d\theta = \frac{\pi}{24}(17^{3/2} - 1) \].

Step 11: Final result

The final answer is the evaluated integral which provides the surface area of the portion specified.

Key Concepts

Surface Area Calculation

When calculating the surface area of a surface defined by a function, it's important to understand the basics of setting up the integral. In this context, the function in question is often denoted as \( z = f(x, y) \). The surface area \( A \) is then obtained by integrating over a specific domain \( D \).
You use the formula:

• \( A = \iint_{D} \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dA \) to calculate the surface area.
• This formula accounts for the contribution of the gradient slope components in the calculation of the surface area.

With this setup, we adjust the computation of surface area according to the limits provided by the domain \( D \), which in the original problem involves a section limited by a cylinder in the 3D plane.

Polar Coordinates

To tackle complex integrals or problems like this one, converting Cartesian coordinates (\(x\)-\(y\) plane) into polar coordinates can be very useful.
Polar coordinates, described by \( (r, \theta) \), offer a transformation where:

• \( x = r\cos(\theta) \)
• \( y = r\sin(\theta) \)

This method transforms the problem into a simpler integral when we work within circular or radial boundaries, such as a cylinder. In our initial problem, because the domain was bound within a cylinder described by \( x^2 + y^2 = 4 \), it became easier to switch to polar forms, where:

• \( 0 \leq \theta \leq \frac{\pi}{2} \) to account for the first octant
• \( 0 \leq r \leq 2 \) limits the radial distance matching the cylinder radius

Partial Derivatives

Partial derivatives aid in determining the slope of a surface in multivariable calculus. For a function \( z = f(x, y) \), the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) measure how \( z \) changes as \( x \) or \( y \) change independently.

• When you compute \( \frac{\partial z}{\partial x} \) for \( z = x^2 - y^2 \), you get \( 2x \), signifying change along the x-axis.
• Similarly, \( \frac{\partial z}{\partial y} = -2y \), signifies change along the y-axis.

These derivatives are crucial when determining how much each variable (x or y) contributes to the change in the surface. In the calculation of surface area, they factor into the integral to account for the increase in arc length.

Double Integral

The double integral is a concept in multivariable calculus used to integrate over two dimensions, providing a way to calculate quantities over a region such as volumes or surface areas in 3D space.
It is denoted as \( \iint \) and often involves integrating over a specific region \( D \) of the \( xy \)-plane.

• For surface area calculations, the double integral surrounds the expression derived from partial derivatives like \( \sqrt{1 + 4r^2} \cdot r \), after converting into polar coordinates.
• In the given problem, the setup was \( \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \sqrt{1 + 4r^2} \cdot r \, dr \, d\theta \).

By evaluating this setup first over \( r \) and subsequently \( \theta \), we find the contribution of the entire bound region in polar form, accumulating the individual sub-regions' surface areas into a total.