Problem 16 - APEX Calculus

Question

Find the mass/weight of the lamina described by the region \(R\) in the plane and its density function \(\delta(x, y)\). \(R\) is the circle sector bounded by \(x^{2}+y^{2}=25\) in the first quadrant; \(\delta(x, y)=\left(\sqrt{x^{2}+y^{2}}+1\right) \mathrm{kg} / \mathrm{m}^{2}\)

Step-by-Step Solution

Answer

The mass of the lamina is \(\frac{325\pi}{12} \text{ kg}\).

Step 1: Identify the Region of Integration

The region, denoted by \(R\), is a sector in the first quadrant of a circle with the equation \(x^2 + y^2 = 25\). Since it is a circle centered at the origin with radius 5, and only in the first quadrant, the angles range from \(0\) to \(\frac{\pi}{2}\).

Step 2: Set Up the Double Integral

To find the mass of the lamina with density function \(\delta(x, y)\), we use the double integral over the region \(R\). The density function \(\delta(x, y)\) is given by \(\delta(x, y) = \left(\sqrt{x^2 + y^2} + 1\right)\). The mass \(M\) is calculated by \(M = \int\int_R \delta(x, y) \, dA\).

Step 3: Convert to Polar Coordinates

Since \(R\) is a circular sector, it is convenient to use polar coordinates:\(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). The density function becomes \(\delta(r, \theta) = r + 1\) since \(\sqrt{x^2+y^2} = r\). The differential area element \(dA\) becomes \(r \, dr \, d\theta\).

Step 4: Define the Limits for Polar Coordinates

For the first quadrant, \(\theta\) ranges from \(0\) to \(\frac{\pi}{2}\), and \(r\) ranges from \(0\) to \(5\) (the radius of the circle).

Step 5: Evaluate the Double Integral

Now, evaluate the integral using polar coordinates:\[M = \int_{0}^{\frac{\pi}{2}} \int_{0}^{5} (r + 1) \, r \, dr \, d\theta\]Split the integral as follows:\[M = \int_{0}^{\frac{\pi}{2}} \int_{0}^{5} (r^2 + r) \, dr \, d\theta\]First solve the inner integral:\[\int_{0}^{5} (r^2 + r) \, dr = \left[\frac{r^3}{3} + \frac{r^2}{2} \right]_{0}^{5} = \left(\frac{125}{3} + \frac{25}{2}\right)\]\[= \frac{250}{6} + \frac{75}{6} = \frac{325}{6}\]Now solve the outer integral:\[M = \int_{0}^{\frac{\pi}{2}} \frac{325}{6} \, d\theta = \frac{325}{6} \left[ \theta \right]_{0}^{\frac{\pi}{2}} = \frac{325}{6} \times \frac{\pi}{2} = \frac{325\pi}{12}\]The mass of the lamina is \(\frac{325\pi}{12} \text{ kg}\).

Step 6: State the Final Answer

The mass of the lamina, which is the weight given the density function, is \(\frac{325\pi}{12} \text{ kg}\) within the specified region.

Key Concepts

Mass Calculation

The concept of mass calculation ensures we can find the weight of an object based on its size and how much material it contains. When dealing with objects of varying density, as in this problem, it becomes a bit more intricate. In physics and calculus, the mass of an object can be computed using integrals. This requires us to integrate the density over the entire volume or, in this case, area of the object.

In our circle sector problem, we use a double integral to calculate mass. We do this because we are working over a two-dimensional region, which is the circular sector. The double integral takes into account the density of every infinitesimally small piece of the region. Therefore, it's essentially adding up the mass of each tiny piece to get the total. This is how we find that the mass of our particular lamina is \( \frac{325\pi}{12} \text{ kg} \).

Density Function

A density function describes how mass is distributed over a region. In simpler terms, it tells you how much material is present per unit area. Density is crucial when an object does not have uniform mass distribution.

For our exercise, the given density function is \( \delta(x, y) = \left(\sqrt{x^2+y^2} + 1\right) \). This function shows that the density depends on the distance from the origin, plus a constant. Such a function implies that the further away you are from the center, the denser the material becomes. To perform the calculations easily and accurately, we transformed this density function into its polar equivalent \( \delta(r, \theta) = r + 1 \), where \( r \) is the radial distance from the origin.

Polar Coordinates

When dealing with circular or sector shaped regions, like in our circle sector, polar coordinates simplify the process. This system uses a radius and angle to specify a point in the plane, rather than the typical x and y Cartesian coordinates.

In polar coordinates, any point is defined by two values:

• \( r \): The distance from the origin.
• \( \theta \): The angle with respect to the positive x-axis.

For our problem, the transformation is \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). This makes integrating over the circular sector more straightforward than using Cartesian coordinates.

Double Integral

A double integral extends the concept of an integral to two dimensions. It is used to compute values over a region in the plane, such as area, volume, or in this case, mass.

For our circle sector challenge, the double integral takes the form \( \int \int_R \, \delta(x, y) \, dA \), with \( dA \) as the infinitesimal area element. In polar coordinates, \( dA \) changes to \( r \, dr \, d\theta \), to match our region's geometry. This represents all the tiny pieces over the circular sector being summed up.

By first solving the inner integral with respect to \( r \) and then the outer integral with respect to \( \theta \), we determine the total mass of the lamina. This approach shows the power of calculus in solving complex geometric problems.