Question

Suppose a thin rectangular plate, represented by a region \(R\) in the \(x y\) -plane, has a density given by the function \(\rho(x, y) ;\) this function gives the area density in units such as grams per square centimeter \(\left(\mathrm{g} / \mathrm{cm}^{2}\right)\) The mass of the plate is \(\iint_{R} \rho(x, y) d A .\) Assume that \(R=\\{(x, y): 0 \leq x \leq \pi / 2,0 \leq y \leq \pi\\}\) and find the mass of the plates with the following density functions. a. \(\rho(x, y)=1+\sin x\) b. \(\rho(x, y)=1+\sin y\) c. \(\rho(x, y)=1+\sin x \sin y\)

Short answer

Question: Find the mass of the plates with the density functions (a) $\rho(x, y) = 1 + \sin x$, (b) $\rho(x, y) = 1 + \sin y$, and (c) $\rho(x, y) = 1 + \sin x \sin y$, where the plate is a thin rectangular plate in the $xy$-plane and the region $R$ is defined by $0 \leq x \leq \pi / 2$ and $0 \leq y \leq \pi$. Answer: The mass of the plates with density functions (a) $\rho(x, y) = 1 + \sin x$, (b) $\rho(x, y) = 1 + \sin y$, and (c) $\rho(x, y) = 1 + \sin x \sin y$ are $\boxed{\pi^2/2}$, $\boxed{\pi^2/2}$, and $\boxed{\frac{3}{4}\pi^2 + \frac{\pi}{4}}$, respectively.

Step-by-step solution

Density Function a: \(\rho(x, y) = 1 + \sin x\)

To find the mass of the plate with density function \(\rho(x, y) = 1 + \sin x\), we need to compute the double integral \(\iint_{R} (1 + \sin x) dA\). Since we know the bounds of the region \(R\), the double integral is: $$ \int_{0}^{\pi} \int_{0}^{\pi/2} (1 + \sin x) dy\,dx $$ First, we integrate with respect to \(y\): $$ \int_{0}^{\pi/2} [(1 + \sin x)y]_{0}^{\pi} dx $$ Expanding the integral, we get: $$ \int_{0}^{\pi/2}(\pi + \pi\sin x) dx $$ Now, we integrate with respect to \(x\): $$ [(\pi x - \pi\cos x)]_{0}^{\pi/2} $$ Plugging the bounds of the integral, the mass of the plate is: $$ \pi(\pi/2) - \pi - (\pi(0) - \pi) $$ The mass of the plate with density function \(\rho(x, y) = 1 + \sin x\) is \(\boxed{\pi^2/2}\).

Density Function b: \(\rho(x, y) = 1 + \sin y\)

To find the mass of the plate with density function \(\rho(x, y) = 1 + \sin y\), we need to compute the double integral \(\iint_{R} (1 + \sin y) dA\). The double integral is: $$ \int_{0}^{\pi/2} \int_{0}^{\pi} (1 + \sin y) dx\,dy $$ First, we integrate with respect to \(x\): $$ \int_{0}^{\pi} [(1 + \sin y)x]_{0}^{\pi/2} dy $$ Expanding the integral, we get: $$ \int_{0}^{\pi}(\frac{\pi}{2} + \frac{\pi}{2}\sin y) dy $$ Now, we integrate with respect to \(y\): $$ [(\frac{\pi}{2}y - \frac{\pi}{2}\cos y)]_{0}^{\pi} $$ Plugging the bounds of the integral, the mass of the plate is: $$ \frac{\pi^2}{2} - \frac{\pi}{2} - (\frac{0}{2} - \frac{\pi}{2}) $$ The mass of the plate with density function \(\rho(x, y) = 1 + \sin y\) is \(\boxed{\pi^2/2}\).

Density Function c: \(\rho(x, y) = 1 + \sin x \sin y\)

To find the mass of the plate with density function \(\rho(x, y) = 1 + \sin x \sin y\), we need to compute the double integral \(\iint_{R} (1 + \sin x \sin y)dA\). The double integral is: $$ \int_{0}^{\pi} \int_{0}^{\pi/2} (1 + \sin x \sin y) dy\,dx $$ First, we integrate with respect to \(y\): $$ \int_{0}^{\pi} [(y - \cos x\cos y)]_{0}^{\pi/2} dx $$ Expanding the integral, we get: $$ \int_{0}^{\pi}(\frac{\pi}{2} - \frac{\cos x}{2}) dx $$ Now, we integrate with respect to \(x\): $$ [(\frac{\pi x}{2} - \frac{1}{4}\sin x)]_{0}^{\pi} $$ Plugging the bounds of the integral, the mass of the plate is: $$ \frac{\pi^2}{2} + \frac{\pi}{4} - (\frac{0}{2} + \frac{1}{4}) $$ The mass of the plate with density function \(\rho(x, y) = 1 + \sin x \sin y\) is \(\boxed{\frac{3}{4}\pi^2 + \frac{\pi}{4}}\).

Key concepts

Density Function

The density function, often denoted by \( \rho(x, y) \), is essential in determining how mass is distributed across a given area. In the context of a thin plate in the plane, this function provides a numeric value at each point \((x, y)\). This value represents the density at that specific point, typically in units like grams per square centimeter (g/cm²). Understanding the density function is integral to calculating properties like mass. It essentially tells us how 'heavy' each tiny piece of the plate is, depending on its location. For a problem involving a thin plate with a given density function, decoding \( \rho(x, y) \) properly allows us to use it in further calculations, such as finding the total mass of the plate. For each exercise, different expressions for the density function are provided, such as \( \rho(x, y) = 1 + \sin x \). Each variation affects how mass is calculated in the respective scenarios.

Mass of a Plate

The mass of a plate can be understood by visualizing the plate divided into infinitesimally small pieces, with each having its unique density given by \( \rho(x, y) \). To find the total mass of the plate, we must sum up the masses of all these infinitesimal pieces. This is achieved through the mathematical operation known as double integration. For the given exercise, the mass \( M \) of the plate is determined using the formula:\[ M = \iint_{R} \rho(x, y) \, dA \]Here, \( dA \) represents a tiny element of area over which the density is averaged, and \( R \) is the region of interest, which in this case is a rectangle defined by the bounds \( 0 \leq x \leq \frac{\pi}{2} \) and \( 0 \leq y \leq \pi \). By solving this integral, we combine all the tiny portions of the mass across the whole plate, accounting for the specific density at each point.Thus, calculating the integral involves two main steps:

• First, integrating with respect to one variable to consider changes in that direction,
• Then with respect to the other, to sweep across the entire region.

Area Integral

Area integrals, like the double integral used to compute the mass of a plate, are a powerful mathematical tool that allows us to sum up quantities distributed over a region in space. In these exercises, we deal with integrals over the rectangle \( R \), bounded in the \( x \)-direction from 0 to \( \frac{\pi}{2} \) and in the \( y \)-direction from 0 to \( \pi \).The expression \( \iint_{R} \rho(x, y) \, dA \) is a perfect illustration of an area integral. This setup helps us explore the cumulative effect of density over a given area.\To solve such integrals, we initially choose one variable to integrate with respect to, while treating the other as a constant. After evaluating this first integration, we move on to the second variable. Typically, the order of integration depends on the given limits, aiming to simplify the process whenever possible.Thus, area integrals not only help us find mass but are also useful in various applications where we need to accumulate quantities spread over an area, such as finding averages, centroids, and more in physical contexts.