Problem 5 - Advanced Engineering Mathematics

Question

Find the surface area of that portion of the paraboloid \(z=4-x^{2}-y^{2}\) that is above the \(x y\) -plane.

Step-by-Step Solution

Answer

The surface area is \(\frac{\pi}{6} (17\sqrt{17} - 1)\).

Step 1: Understand the problem

We need to find the surface area of the portion of the paraboloid defined by the equation \( z = 4 - x^2 - y^2 \) that is above the \(xy\)-plane. The paraboloid opens downwards, and we must ensure that \(z \geq 0\).

Step 2: Determine the region of the paraboloid above the xy-plane

Since \(z \geq 0\), set \(4 - x^2 - y^2 \geq 0\). This simplifies to \(x^2 + y^2 \leq 4\), which describes a circle centered at the origin with radius 2, in the \(xy\)-plane.

Step 3: Parameterize the surface

Parameterize the surface using cylindrical coordinates: Let \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z = 4 - r^2\) where \(0 \leq r \leq 2\) and \(0 \leq \theta < 2\pi\).

Step 4: Calculate the surface area using the formula

The surface area \(A\) can be found using the surface integral: \[ A = \iint_{D} \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \ dA \]Calculate the partial derivatives: \(\frac{\partial z}{\partial x} = -2x\) and \(\frac{\partial z}{\partial y} = -2y\).

Step 5: Evaluate the integrand

Substitute \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) into the formula: \(\sqrt{1 + 4x^2 + 4y^2} = \sqrt{1 + 4(x^2 + y^2)}\). In cylindrical coordinates, this becomes \(\sqrt{1 + 4r^2}\).

Step 6: Set up the integral

The integral for the surface area becomes: \[ A = \int_{0}^{2\pi} \int_{0}^{2} \sqrt{1 + 4r^2} \cdot r \ dr \ d\theta\]

Step 7: Solve the integral

First, evaluate the inner integral with respect to \(r\):\[ \int_{0}^{2} r\sqrt{1 + 4r^2} \ dr\]Use substitution \(u = 1 + 4r^2\), \(du = 8r\ dr\), or \(r\ dr = \frac{1}{8} du\). When \(r = 0, u = 1\) and when \(r = 2, u = 17\). Therefore,\[ \frac{1}{8} \int_{1}^{17} u^{1/2} du\]which results in \(\frac{1}{8} \times \frac{2}{3} \times (17^{3/2} - 1^{3/2}) = \frac{1}{12} (17\sqrt{17} - 1)\).

Step 8: Integrate over \(\theta\)

Now, integrate over \(\theta\):\[ A = \int_{0}^{2\pi} \frac{1}{12} (17\sqrt{17} - 1) \ d\theta = \frac{1}{12} (17\sqrt{17} - 1) \times 2\pi\]

Step 9: Simplify the final expression

Show the result:\[ A = \frac{\pi}{6} (17\sqrt{17} - 1) \]

Key Concepts

Surface Integrals

Surface integrals are an essential concept in multivariable calculus and provide us with a method to calculate the area of a surface embedded in three-dimensional space. Imagine you want to measure the area of a ripple on water. Unlike flat surfaces, these areas require more intricate calculations.
To find a surface area, we use a surface integral, which is essentially an extension of line integrals you might have encountered for curves.For the paraboloid exercise, the surface integral involves calculating an integral over a parameterized surface. This surface is determined by a function like our paraboloid, defined as \(z=4-x^2-y^2\). The surface integral formula involves a specific term:

• \( \iint_{D} \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \ dA \)

This expression accounts for how sloped the surface is at each point, derived from partial derivatives of \(z\). It tells us how the surface stretches across the \(xy\)-plane and the additional area contributed by its elevation.Calculating this involves integrating a function which usually requires a lot of substitutions, as seen in the problem's solution. To get an accurate answer, every part of the paraboloid must be considered, wrapped up through an integral over its entire domain, \(D\), which in cylindrical coordinates is a disk from \(r=0\) to \(2\) and \(\theta=0\) to \(2\pi\).

Cylindrical Coordinates

Cylindrical coordinates are a type of coordinate system that simplifies the process of finding surface areas and volumes of objects with circular symmetry. They are incredibly useful, especially for surfaces like paraboloids, cylinders, and cones that naturally have rotational symmetry around an axis.
Cylindrical coordinates work by expressing positions in the plane with a radius and angle, akin to using polar coordinates, but extended to three dimensions by including height, \(z\).

• The conversion for this system includes: \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z\) is often given as a function of \(x\) and \(y\) (or \(r\)).
• The range for \(r\) is from the origin outwards, \(\theta\) is the angle from a reference direction, and \(z\) measures vertical displacement.

In the problem at hand, by recognizing the circular nature of the paraboloid cap, using cylindrical coordinates makes the integration process much simpler.The region described, \(x^2 + y^2 \leq 4\), is a circle of radius 2 centered at the origin in the \(xy\)-plane. Transitioning to cylindrical coordinates allows us to handle this region with more straightforward integrals over \(r\) and \(\theta\). The parameterization is very effective here because it leverages symmetry which is not as apparent in Cartesian coordinates.

Partial Derivatives

Partial derivatives are a crucial tool in calculus that quantify how a multivariable function changes as one of its independent variables changes while keeping others constant. They are like the digital zoom of calculus—enabling a focused view on how a function's shape morphs with a single variable at a time.
For surfaces like our paraboloid, partial derivatives tell us how sloped the surface is in the direction of the \(x\) and \(y\) axes. They play a significant role in determining the surface's behavior.

• For a surface \(z=f(x,y)\), the partial derivative \(\frac{\partial z}{\partial x}\) measures the rate of change of \(z\) in the \(x\) direction.
• Similarly, \(\frac{\partial z}{\partial y}\) represents the rate of change in the \(y\) direction.

In our example, we calculated these derivatives for \(z = 4 - x^2 - y^2\):

• \(\frac{\partial z}{\partial x} = -2x\)
• \(\frac{\partial z}{\partial y} = -2y\)

These derivatives help us compute how much extra area the tilt of the surface contributes to the basic \(xy\)-plane area. The steeper these slopes, the more surface area we have to account for—like how more paper is needed to cover a hill than flat ground. In integral terms, they are part of the expression that adjusts for the three-dimensional stretch of the surface.