Problem 55 - Calculus Early Transcendentals

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. \(\overline{\rho(x, y)}=1+\sin x \sin y\)

Step-by-Step Solution

Answer

Question: Calculate the mass of thin rectangular plates with the given density functions over the region \(0 \leq x \leq \frac{\pi}{2}\), \(0 \leq y \leq \pi\): a. \(\rho(x, y) = 1 + \sin x\) b. \(\rho(x, y) = 1 + \sin y\) c. \(\rho(x, y) = 1 + \sin x \sin y\) Answer: a. \(Mass_a = 2\pi\) (grams) b. \(Mass_b = \pi^2\) (grams) c. \(Mass_c = -\pi^2\) (grams)

Step 1: Calculate the mass for density function a.

Set up the double integral for density function a: \(\rho(x, y) = 1 + \sin x\). $$Mass_a = \iint_{R} (1+\sin x) dA$$ Now we need to integrate over the region R. First, we will integrate with respect to x and then with respect to y: $$Mass_a = \int_0^{\pi}\int_0^{\frac{\pi}{2}} (1+\sin x) dxdy$$

Step 2: Calculate the mass for density function b.

Set up the double integral for density function b: \(\rho(x, y) = 1 + \sin y\). $$Mass_b = \iint_{R} (1+\sin y) dA$$ Now we need to integrate over the region R. First, we will integrate with respect to x and then with respect to y: $$Mass_b = \int_0^{\pi}\int_0^{\frac{\pi}{2}} (1+\sin y) dxdy$$

Step 3: Calculate the mass for density function c.

Set up the double integral for density function c: \(\rho(x, y) = 1 + \sin x \sin y\). $$Mass_c = \iint_{R} (1+\sin x \sin y) dA$$ Now we need to integrate over the region R. First, we will integrate with respect to x and then with respect to y: $$Mass_c = \int_0^{\pi}\int_0^{\frac{\pi}{2}} (1+\sin x \sin y) dxdy$$

Step 4: Evaluate the integrals and find the mass of the plates.

We will now evaluate the three integrals: For density function a: $$Mass_a = \int_0^{\pi}\int_0^{\frac{\pi}{2}} (1+\sin x) dxdy = \int_0^{\pi} [\frac{\pi}{2} - \cos x]_0^{\frac{\pi}{2}} dy = \int_0^{\pi} (2) dy = 2\pi$$ For density function b: $$Mass_b = \int_0^{\pi}\int_0^{\frac{\pi}{2}} (1+\sin y) dxdy = \int_0^{\pi} [\frac{\pi}{2} + (\frac{\pi}{2} -1)\cos y]_0^{\pi} dx = \int_0^{\pi} (\pi - \cos y) dx = \pi^2$$ For density function c: $$Mass_c = \int_0^{\pi}\int_0^{\frac{\pi}{2}} (1+\sin x \sin y) dxdy$$ First, integrate with respect to x: $$ = \int_0^{\pi} [\frac{\pi}{2} - \cos{x} \sin{y} ]_0^{\frac{\pi}{2}} dy$$ Now, integrate with respect to y: $$ = \int_0^{\pi} [2 - (\pi - \cos{y})] dy = \int_0^{\pi} [2 - \pi + \cos{y}] dy$$ $$ = \int_0^{\pi} [(\cos{y} - \pi) + 2] dy = \int_0^{\pi} [(\cos{y} + 1) - \pi] dy$$ $$ = [\sin y - \pi y]_0^{\pi} = (\sin{\pi} - \pi^2) - (\sin0 - 0) = -\pi^2$$ So the mass of the plates with the given density functions is: a. \(Mass_a = 2\pi\) (grams) b. \(Mass_b = \pi^2\) (grams) c. \(Mass_c = -\pi^2\) (grams)

Key Concepts

Density Function

A density function tells us how mass is spread across a region. It's like a map of how heavy or light different parts of an object are. For a thin rectangular plate in the xy-plane, the density function \(\rho(x, y)\) represents mass per area unit, such as grams per square centimeter (g/cm²). This density function can depend on the location in the plate, sometimes influenced by physical properties such as material or temperature.

Think of this density function as a mathematical thermometer of mass. If you have \(\rho(x, y) = 1 + \sin x\), it means that at each point \((x, y)\) on the plate, the local density consists of a base value of 1 plus some variation tied to the sine of the \(x\)-coordinate. Depending on this formula, the realm of heavier or lighter areas can shift, making some sections denser than others.

Mass Calculation

Finding the mass of a plate with a non-uniform density involves using double integrals. This is because the density varies across the plate, and we need to add up tiny pieces of mass over the entire region \(R\). To compute the total mass, we integrate the density function \(\rho(x, y)\) over the plate's area.

• First, define the region \(R\) over which to integrate. For this exercise, \(R = \{(x, y): 0 \leq x \leq \pi/2, 0 \leq y \leq \pi\}\).
• Next, write the double integral \(Mass = \iint_R \rho(x, y) \, dA\). In this form, \(dA\) represents an infinitesimal piece of area within the region \(R\).
• For each specific density, such as \(\rho(x, y) = 1 + \sin x\), plug the density expression into the integral and solve step by step.

This computation helps to convert a problem of varying mass into a single integral, making it solvable with calculus techniques.

Rectangular Plate

When working with rectangular plates, it's vital to understand their geometry for the integral setup. Here, the rectangle is defined by the bounds \(0 \leq x \leq \pi/2\) and \(0 \leq y \leq \pi\). These bounds form a region, \(R\), and this is where the magic of integration takes place.

A rectangular plate is like a sheet, vast and uniform, yet the density can vary. We see this in problems with different density functions, each telling a story of how the mass is distributed. Integration over such plates involves:

• Setting up limits of integration: First over \(x\) from 0 to \(\pi/2\), then over \(y\) from 0 to \(\pi\).
• Evaluating integrals: Performing calculations with care to capture how mass piles up across length and width.

By understanding how to incorporate the geometry and density together, we can effectively tackle the problem of finding mass distributed along a rectangular plate.