Question

Find the center of mass of the region in the first quadrant bounded by the circle \(x^{2}+y^{2}=a^{2}\) and the lines \(x=a\) and \(y=a,\) where \(a > 0\).

Short answer

Answer: The coordinates of the center of mass of the given region are \(\left(\frac{2a}{3\pi}, \frac{a}{3}\right)\).

Step-by-step solution

Set up the integral for the area of the region.

First, let's find the area of the region. To do this, we need to express y as a function of x for the curve, which is given by the equation \(x^2+y^2=a^2\). Solving for y in terms of x, we get \(y(x)=\sqrt{a^2-x^2}\). Now, we can set up the integral for the area, which will be the integral of the curve, subtracting the integral of the straight line x=a. The Area \(A\) of the region can be given by: \(A=\int_{0}^{a} (\sqrt{a^2-x^2} - 0) dx.\)

Calculate the area of the region.

Now, let's compute the area of the region using the integral we set up in the previous step: \(A=\int_{0}^{a} \sqrt{a^2-x^2} dx\) This integral can be solved using the trigonometric substitution: \(x = a\sin{\theta}\), \(dx = a\cos{\theta}d\theta\). The new limits of integration will be \(0\) and \(\frac{\pi}{4}\). Now substitute to get: \(A=a\int_{0}^{\frac{\pi}{4}} \sqrt{a^2 - a^2\sin^2{\theta}} \cos{\theta} d\theta\) \(A=a\int_{0}^{\frac{\pi}{4}} a\cos^2{\theta} d\theta\) \(A=a^2 \int_{0}^{\frac{\pi}{4}} \cos^2{\theta} d\theta\) Now we can solve the integral using the half-angle identity \(\cos^2{\theta} = \frac{1+\cos{2\theta}}{2}\): \(A=a^2\int_{0}^{\frac{\pi}{4}} \frac{1+\cos{2\theta}}{2} d\theta\) \(A=a^2\left[\frac{1}{2}\theta+\frac{1}{4}\sin{2\theta}\right]_{0}^{\frac{\pi}{4}}\) \(A=\frac{a^2\pi}{4}\)

Set up the integrals for the moments with respect to x and y axes.

Now let's find the moment with respect to the x and y axes. Moment with respect to x-axis: \(\bar{y}=\frac{1}{A}\int_{0}^{a} x\sqrt{a^2-x^2} dx.\) Moment with respect to y-axis: \(\bar{x}=\frac{1}{A}\int_{0}^{a} y(a-y) dy.\)

Compute the moments with respect to x and y axes.

Moment with respect to x-axis: Similar to step 2, let's use the trigonometric substitution \(x=a\sin{\theta}\). Then, the moment with respect to the x-axis is given by: \(\bar{y}=\frac{1}{\frac{a^2\pi}{4}}\int_{0}^{\frac{\pi}{4}} a\sin{\theta}\cdot a\sqrt{a^2-a^2\sin^2{\theta}} a\cos{\theta} d\theta\) \(\bar{y}=\frac{4}{\pi}\int_{0}^{\frac{\pi}{4}}\sin{\theta}\cdot a^2 \cos^2{\theta} d\theta\) Now, we can use our previously determined value for the Area and solve the integral as before to get: \(\bar{y} = \frac{a}{\pi}\int_{0}^{\frac{\pi}{4}}\sin{\theta}\cdot(1 + \cos{2\theta}) d\theta\) \(\bar{y} = \frac{a}{\pi}\int_{0}^{\frac{\pi}{4}}\sin{\theta} + \sin{\theta}\cdot\cos{2\theta} d\theta\) \(\bar{y} = \frac{a}{\pi}\left[-\cos{\theta}\left|_{0}^{\frac{\pi}{4}}\right. +\frac{1}{3}\cos^3{\theta}\left|_{0}^{\frac{\pi}{4}}\right.\right]\) \(\bar{y} = \frac{a}{3}\) Moment with respect to y-axis: For the moment with respect to the y-axis, we need to look at the rectangular part and the triangular part of the region separately. \(M_x = \int_{0}^{a}y(a-y)dy = \int_{0}^{a} ay - y^2 dy \) \(\bar{x} = \frac{1}{\frac{a^2\pi}{4}}\cdot M_x \) Now we compute the moment: \(M_x = \left[ \frac{1}{2}ay^2 - \frac{1}{3}y^3\right]_0^a = \frac{1}{2}a^3 - \frac{1}{3}a^3= \frac{1}{6}a^3\) And now compute the center of mass coordinate \(\bar{x}\): \(\bar{x} = \frac{4}{\pi}\cdot\frac{1}{6}a = \frac{2a}{3\pi}\)

Write the center of mass coordinates using the moments computed in Step 4.

Now that we have computed the moments with respect to x and y axes, we can write the coordinates of the center of mass as (x_c, y_c) = (\(\bar{x}\), \(\bar{y}\)): \(\text{Center of Mass} = \left(\frac{2a}{3\pi}, \frac{a}{3}\right)\)

Key concepts

Trigonometric Substitution

Trigonometric substitution is a useful technique in integral calculus. It involves substituting a trigonometric function for a variable to simplify the integration process.
In our problem, we use trigonometric substitution to solve the integral \(\int \sqrt{a^2-x^2} \, dx\). This type of integral, which contains a square root of a squared variable minus another squared variable, is a classic case for trigonometric substitution.
The typical substitution here is \( x = a\sin{\theta} \). Consequently, \( dx = a\cos{\theta} \, d\theta \). By replacing these values in the integral, we can eventually simplify it to easier terms that can be integrated using basic trigonometric identities.
In this way, trigonometric substitution helps in transforming a potentially complex integral into a more manageable form, allowing us to ultimately find the area under the curve or region more easily.

Integral Calculus

Integral calculus is fundamentally about finding the total or accumulative value; in this case, the integral helps find the area of a region.
Using integral calculus, we can compute areas between curves, spaces between boundaries, and other related measurements by summing infinitely small quantities.
We start with the function \( y(x) = \sqrt{a^2-x^2} \), derived from the circle equation \( x^2 + y^2 = a^2 \). The aim is to determine the "area" between the curve and the coordinate axes in the first quadrant, just up to the boundaries of \( x = a \) and \( y = a \).
The integral calculus approach involves setting up a proper definite integral, in our case \( \int_0^a \sqrt{a^2-x^2} \, dx \), to find the precise area of this region. This integral computes the sum of infinitesimally small segments of the area, leading to an exact solution for computations such as the center of mass.

Moments of Inertia

'Moments' in this context refer to rotational properties about an axis. The moment of a shape about a particular axis is essentially that shape's tendency to rotate around that axis.
In our exercise, the moment calculations help us derive coordinates for the center of mass (\( \bar{x} \) and \( \bar{y} \)). Specifically: - **Moment with respect to the x-axis** involves the formula \( \bar{y} = \frac{1}{A} \int x \sqrt{a^2-x^2} \, dx \), capturing how the mass is distributed vertically. - **Moment with respect to the y-axis** involves \( \bar{x} = \frac{1}{A} \int y(a-y) \, dy \), illustrating how the mass spreads horizontally.
Calculating these moments provides insights into how the mass or area of a shape would behave under constraints of rotation, pivotal for locating the precise center of mass.

Definite Integrals

Definite integrals are a key concept in calculus, providing the accumulated value of a function over a specified interval.
They are critical in calculating actual numerical values for lengths, areas, volumes, and other quantities. In our problem, definite integrals allow us to determine the exact area of the region in the first quadrant bounded by the given circle and the lines \( x = a \) and \( y = a \).
By evaluating \( \int_{0}^{a} \sqrt{a^2-x^2} \, dx \), we move from abstract equations to concrete mathematical solutions, such as the resulting area being \( \frac{a^2\pi}{4} \).
Furthermore, these integrals play a significant role when we find the center of mass, enabling precise calculations by quantifying interactions between points on the curved surface and the system's boundaries.