Problem 33 - University Calculus: Early Transcendentals

Question

In polar coordinates, the average value of a function over a region \(R\) (Section 14.3 ) is given by $$\frac{1}{\operatorname{Area}(R)} \iint_{R} f(r, \theta) r d r d \theta$$ Find the average height of the hemispherical surface \(z=\sqrt{a^{2}-x^{2}-y^{2}}\) above the disk \(x^{2}+y^{2} \leq a^{2}\) in the \(x y\)-plane.

Step-by-Step Solution

Answer

The average height is \(\frac{2}{3}a\).

Step 1: Convert to Polar Coordinates

We start by converting the Cartesian coordinates to polar coordinates. For a disc defined by \(x^2 + y^2 \leq a^2\), we utilize the relationships \(x = r \cos\theta\) and \(y = r \sin\theta\). Therefore, \(x^2 + y^2 = r^2\) and the equation becomes \(r \leq a\).

Step 2: Define the Function in Polar Coordinates

The surface is defined by \(z = \sqrt{a^2 - x^2 - y^2}\). Substituting \(x\) and \(y\) in terms of \(r\) and \(\theta\), we get \(z = \sqrt{a^2 - r^2}\). Thus, the function \(f(r, \theta) = \sqrt{a^2 - r^2}\).

Step 3: Set Up the Double Integral

To find the average height, we need to evaluate \(\frac{1}{\text{Area}(R)} \iint_R f(r, \theta) r \, dr \, d\theta\). Here, \(R\) is the region where \(0 \leq r \leq a\) and \(0 \leq \theta \leq 2\pi\). Therefore, the integral becomes \(\iint_R \sqrt{a^2 - r^2} \, r \, dr \, d\theta\).

Step 4: Compute the Area of the Disc

The area \(\text{Area}(R)\) of the disc given by \(x^2 + y^2 \leq a^2\) is \(\pi a^2\). This is straightforwardly computed using the formula for the area of a circle with radius \(a\).

Step 5: Evaluate the Inner Integral

We first compute the inner integral over \(r\), so we have \(\int_0^a \sqrt{a^2 - r^2} \, r \, dr\). This integral can be solved using the substitution \(r = a \sin\theta\), leading to \(dr = a \cos\theta \, d\theta\) and the bounds \(0\) to \(\frac{\pi}{2}\). Solving the resulting expression gives \(\frac{1}{3}a^3\).

Step 6: Evaluate the Outer Integral

Substitute the result of the inner integral \(\frac{1}{3}a^3\) into the outer integral with respect to \(\theta\): \(\int_0^{2\pi} \frac{1}{3}a^3 \, d\theta = \frac{1}{3}a^3 \cdot 2\pi = \frac{2}{3}\pi a^3\).

Step 7: Compute the Average Value

Now, calculate the average value using the formula \(\frac{1}{\text{Area}(R)} \iint_R f(r, \theta) r \, dr \, d\theta\), which becomes \(\frac{1}{\pi a^2}\cdot \frac{2}{3} \pi a^3\). This simplifies to \(\frac{2}{3}a\).

Key Concepts

Average Value of a Function

The average value of a function over a certain region in polar coordinates provides a useful way to understand the function's overall behavior within that region. This concept is particularly handy in applications involving areas that are circular or have symmetrical properties. To compute this average, we use the formula:

• \(\frac{1}{\operatorname{Area}(R)} \iint_{R} f(r, \theta) r \, dr \, d\theta\)

Here, \(R\) is the region over which we want to find the average, and \(f(r, \theta)\) is our function in polar coordinates. By integrating \(f(r, \theta)\) with respect to \(r\) and \(\theta\), we accumulate its values over the entire region, while the multiplication by \(r\) accounts for the radial nature of the space.
This process effectively "spreads out" the value of the function over the whole region and, when divided by the region's area, gives us a single number that represents the average value of the function on that domain.

Double Integral

Double integrals extend the concept of integration to functions of two variables, allowing us to calculate the volume under surfaces in 3D space. For functions expressed in polar coordinates, a double integral looks like this:

• \(\iint_R f(r, \theta) r \, dr \, d\theta\)

This involves integrating first with respect to the radial coordinate \(r\), and then with respect to the angular coordinate \(\theta\). The presence of the term \(r\) in the product \(r \, dr \, d\theta\) is critical, as it accounts for the fact that small changes in \(r\) cover larger areas the further they are from the origin.
Performing a double integral can be broken into two steps: solving the inner integral with respect to one variable (holding the other constant) and then solving the resulting expression with respect to the other variable. This approach makes it feasible to compute complex areas and volumes, especially useful in physics and engineering applications.

Polar Coordinate Transformation

Polar coordinate transformation is essential when dealing with circular or radially symmetric regions. This transformation is achieved by expressing the Cartesian coordinates \((x, y)\) in terms of polar coordinates \((r, \theta)\), where:

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

The benefit of this transformation is that it simplifies the expression of regions like circles into more manageable boundaries. For example, a circle expressed as \(x^2 + y^2 \leq a^2\) becomes \(r \leq a\) in polar coordinates, which is much simpler to work with when setting integration limits.
Thus, polar coordinate transformation often decreases the complexity of the problems involving circular areas or surfaces, making them more approachable for analysis and integration.

Area of a Disk

The area of a disk in the coordinate plane, especially when expressed in polar coordinates, is straightforward to calculate. Given a disk of radius \(a\), its area \(A\) is computed via the formula:

• \(A = \pi a^2\)

This formula comes from the basic relationship between a circle's radius and its area in Euclidean geometry. In polar coordinates, the disk expressed as \( x^2 + y^2 \leq a^2\) simplifies to \( r \leq a\), and the area integral becomes:

• \(\iint_R r \, dr \, d\theta\)

Here, \(R\) represents the circular region defined by \(0 \leq r \leq a\) and \(0 \leq \theta \leq 2\pi\), which, when integrated, recovers the familiar \(\pi a^2\).
Understanding the area of a disk and the simplification in polar form plays a fundamental role in many fields, such as physics and engineering, where symmetrical shapes frequently appear.