Question

As needed, use a computer to plot graphs and to check values of integrals. (a) Find the area inside the circle \(r=2,\) with \(x>0\) and \(y > 1\). (b) Find the centroid of the area in (a).

Short answer

(a) The area is \( \frac{\pi}{3} - 1 \). (b) The centroid is located at \(x_c = 0\) and \(y_c = \frac{3 \pi}{8 - 3} \).

Step-by-step solution

- Setup the problem

The problem requires the area inside a circle of radius 2, but limited to the region where both x>0 and y>1. The polar equation of the circle is given by: \[ r = 2 \]

- Express the limits in Cartesian coordinates

The given circle in Cartesian coordinates is: \[ x^2 + y^2 = 4 \] The regions are: \( x > 0 \) and \( y > 1 \), which sketch a quarter of the circle plus some additional limitations.

- Find intersection points

To find the bounds of integration, solve for the intersection of the circle with the lines \( y = 1 \) and \( x = 0 \). Substitute \( y = 1 \) into the circle equation: \[ x^2 + 1 = 4 \] \[ x^2 = 3 \] \[ x = \sqrt{3} \] Thus, the bounds for x are from 0 to \( \sqrt{3} \).

- Set up the area integral

To compute the area integral, use the boundaries found: \[ \text{Area} = 2 \times \int_{1}^{2} \int_{0}^{\sqrt{4-y^2}} dx \,dy \] Double the integral because the area x > 0 includes up to y > 1 in symmetry.

- Simplify and evaluate the integral

First execute the inner integral with respect to x: \[ \int_{0}^{\sqrt{4-y^2}} \ dx = \sqrt{4-y^2} \] Now the outer integral: \[ \text{Area} = 2 \int_{1}^{2} \sqrt{4-y^2} \,dy \] Evaluate using substitution.

- Solve the outer integral

Using a trigonometric substitution, let \( y = 2 \, \cos(\theta) \): Convert the integral to \[ 2 \int_{0}^{\arccos(1/2)} 2 \, \sin^2(\theta) \, d\theta = \frac{\pi}{3} - 1 \]

- Calculate the centroid coordinates

The centroid can be calculated using the integral definitions for \( x_c \) and \( y_c \): \[ x_c = \frac{1}{A} \int \int_A x dA \] and \[ y_c = \frac{1}{A} \int \int_A y dA \] Solve these integral parts for x and y separately.

- Solve \( x_c \)

\[ x_c = \frac{1}{A} \int_{1}^{2} \int_0^{\sqrt{4-y^2}} x dx dy = 0 \] (by symmetry)

- Solve \( y_c \)

\[ y_c = \frac{1}{A} \int_{1}^{2} \int_{0}^{\sqrt{4-y^2}} y dx dy = \frac{3 \pi}{8 - 3 \] (using integral and proper limits)

Key concepts

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system where each point is determined by a distance from a reference point and an angle from a reference direction. The reference point is called the pole (analogous to the origin in Cartesian coordinates), and the reference direction is typically the positive x-axis.
In this exercise, we are given a circle of radius 2 in polar form: \[ r = 2 \]The points on the circle are described as all points that are 2 units away from the origin, regardless of the angle. This format is especially useful when dealing with circular shapes or sectors because it simplifies the mathematical expressions involved.
To convert between Cartesian coordinates and polar coordinates:

• The coordinate x is given by \( x = r \, \cos(\theta) \)
• The coordinate y is given by \( y = r \, \sin(\theta) \)

These relationships are useful in switching between the coordinate systems when setting up and solving integrals in polar coordinates.

Area Integral

To find the area inside the circle where both x>0 and y>1, we need to set up an integral that captures these conditions. In Cartesian coordinates, the circle equation is:\[ x^2 + y^2 = 4 \]By solving for intersection points with lines y = 1 and x = 0, the integration limits are obtained.Next, we express the area integral using these Cartesian limits then transform it into polar coordinates for simplification:\[ \text{Area} = 2 \times \int_{1}^{2} \, \int_{0}^{\sqrt{4-y^2}} dx \, dy \]In polar coordinates, the problem becomes much more manageable. The radial distance r ranges from 0 to 2 while the angle \( \theta \) ranges over an appropriate angle.The final converted form gives the area in a symmetrical way and is often easier to evaluate, especially for circular shapes or regions.

Centroid Calculation

The centroid of a shape is its geometric center. To find the centroid of the area derived earlier, we use integral formulas. For any 2D shape, the centroid coordinates \( x_c \) and \( y_c \) are given by:

• \( x_c = \frac{1}{A} \int \int_A x \, dA \)
• \( y_c = \frac{1}{A} \int \int_A y \, dA \)

Here, \( A \) is the area found in the previous section. The integral captures the balance point or the average location of mass if the area has uniform density.
In this exercise, by symmetry, \( x_c \) turns out to be zero. This simplifies our calculation substantially. For \( y_c \), we integrate over the specified limits and divide by the area to get the centroid along the y-axis.It's worth noting that setting up and evaluating these integrals require careful consideration of the limits and often shifts to more manageable coordinate systems like polar coordinates.

Trigonometric Substitution

When dealing with integrals involving square roots, trigonometric substitution can greatly simplify the evaluation. In this exercise, to evaluate the outer integral, we use the substitution:\[ y = 2 \, \cos(\theta) \]This transforms the integral into a trigonometric form, which is more straightforward to solve. The idea is that the trigonometric identities help simplify the expressions involving square roots. For example, \( \sin^2(\theta) \) and \( \cos^2(\theta) \) sum to 1, allowing us to simplify the integrands.After substitution, the integral becomes limited in terms of \( \theta \):\[ 2 \int_{0}^{\arccos(1/2)} 2 \, \sin^2(\theta) \, d\theta \]This is simpler to integrate than the original form. Finally, solving such an integral often involves known results or standard forms in integral calculus, offering a much more straightforward path to the solution.