Question

Use the definition for the average value of a function over a region \(R\) (Section 1 ), \(\bar{f}=\frac{1}{\text { area of } R} \iint_{R} f(x, y) d A\). Find the average value of \(z=a^{2}-x^{2}-y^{2}\) over the region \(R=\left\\{(x, y): x^{2}+y^{2} \leq a^{2}\right\\},\) where \(a>0\)

Short answer

Answer: The average value of the function is \(\frac{a^2}{2}\).

Step-by-step solution

Find the area of the region R

The area of the region R, which is a disc, can be found using the formula for the area of a circle: \(\text{area of R} = \pi \cdot a^2\). Therefore, \(\text{area of R} = \pi a^2\).

Convert the integral to polar coordinates

Since the region R is a disc, it is appropriate to use polar coordinates to evaluate the integral. Recall that \(x = r \cos{\theta}\) and \(y = r \sin{\theta}\). We also need to find the Jacobian \(J\) to make the transformation complete: \(J = \frac{\partial(x, y)}{\partial(r, \theta)}\). After calculation, we find \(J=r\): $$ J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos{\theta} & -r \sin{\theta} \\ \sin{\theta} & r \cos{\theta} \end{vmatrix} = (r\cos^2{\theta}+r\sin^2{\theta})=r. $$ Thus, the integral over R becomes: $$ \iint_{R} f(x, y) dA =\iint_{R} (a^2-x^2-y^2) r dr d\theta. $$ with the limits of integration being \(0\leq r \leq a\) for radius, and \(0\leq\theta\leq2\pi\) for the angle.

Substitute the polar coordinates and evaluate the integral

As we now have the expressions for \(x\), \(y\), and the Jacobian, we can substitute them into the integral: $$ \iint_{R}(a^{2}-x^{2}-y^{2})r drd\theta = \int_{0}^{2\pi}\int_{0}^{a}(a^{2}-r^{2}\cos^{2}{\theta}-r^{2}\sin^{2}{\theta})r drd\theta $$ Now, we can factor \(r^2\) out of the expression for the integrand, and simplify: $$ \int_{0}^{2\pi}\int_{0}^{a}(a^{2} - r^2(\cos^2{\theta}+\sin^2{\theta}))r drd\theta = \int_{0}^{2\pi}\int_{0}^{a}(a^2 - r^2)r drd\theta $$ This integral now becomes separable, and we can evaluate the \(r\) and \(\theta\) integrals separately: $$ \int_{0}^{2\pi}\int_{0}^{a}(a^2 - r^2)r drd\theta = \int_{0}^{2\pi} d\theta \int_{0}^{a}(a^2r - r^3) dr. $$ First, integrating with respect to \(r\): $$ \int_{0}^{a}(a^2r - r^3) dr = \left[\frac{1}{2}a^2r^2-\frac{1}{4}r^4\right]_{0}^{a} = \frac{1}{2}a^{4}-\frac{1}{4}a^{4} = \frac{1}{4}a^{4}. $$ Next, integrating with respect to \(\theta\): $$ \int_{0}^{2\pi} d\theta = 2\pi. $$ Multiplying these results together, we get the value of the integral over R to be: $$ \iint_{R} f(x,y)dA = \frac{1}{4}a^{4}\cdot 2\pi = \frac{1}{2}a^{4}\pi. $$

Calculate the average value of the function

To find the average value of the function over the region R, we use the formula given in the exercise: $$ \bar{f}=\frac{1}{\text{area of R}} \iint_{R} f(x, y) dA = \frac{1}{\pi a^2}\cdot \frac{1}{2}a^{4}\pi= \frac{a^{2}}{2}. $$ So, the average value of the function \(z=a^2-x^2-y^2\) over the region \(R=\{(x, y): x^2+y^2\leq a^2\}\), where \(a>0\), is \(\frac{a^2}{2}\).

Key concepts

Polar coordinates integration

When working with regions like discs in the plane, transitioning from Cartesian to polar coordinates can simplify integration. This is because the symmetry of the circular region aligns neatly with the polar coordinate system.
To convert, we utilize the relationships:

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

With these transformations, the element of area \(dA\) becomes \(r \, dr \, d\theta\). This includes the Jacobian \(r\), which accounts for the change of variables. Such transformations make handling circular regions more manageable. Notably, the limits of integration typically range from \(0\) to \(a\) for \(r\) and \(0\) to \(2\pi\) for \(\theta\) in a full circle.

Jacobian transformation

The Jacobian transformation is a vital concept when changing variables in multiple integrals. It accounts for how area elements scale under the transformation. In converting from Cartesian to polar coordinates, the Jacobian is the determinant of the partial derivatives of \(x\) and \(y\) with respect to \(r\) and \(\theta\).
The Jacobian determinant \(J\) is expressed as:

• \[J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = r \]

This shows that each unit area in polar coordinates becomes \(r\) times that area in Cartesian coordinates. Understanding the Jacobian is crucial for correctly setting up integrals after a change of variables, ensuring accurate evaluations.

Multivariable calculus

Multivariable calculus extends calculus to functions of several variables. It involves partial derivatives and multiple integrals. These concepts are foundational for understanding how functions behave in higher dimensions.
With functions of two variables, such as \(f(x,y)\), integration over a region \(R\) involves computing a double integral: \(\iint_{R} f(x,y) \, dA\).
Factors like the region's shape often necessitate changes in coordinates, such as from Cartesian to polar, to simplify calculations. Mastery of multivariable calculus is essential in fields like physics and engineering, where phenomena depend on two or more variables.

Disc integration area

Disc integration focuses on computing integrals over circular regions, commonly using polar coordinates to simplify the process. The area of a disc with radius \(a\) is given by \(\pi a^2\).
When finding the average value of a function over a disc, the concept utilizes the formula:

• \[\bar{f} = \frac{1}{\text{area of } R} \iint_{R} f(x, y) \, dA\]

This simplifies problems involving circular symmetry. By leveraging the disc's geometry and symmetry, such integrations become more feasible. The average value computed gives insights into the function’s behavior across the entire region.