Question

A thin plate of unit density occupies the region between the parabola \(y=a x^{2}\) and the horizontal line \(y=b\) where \(a > 0\) and \(b > 0 .\) Show that the center of mass is \(\left(0, \frac{3 b}{5}\right),\) independent of \(a\)

Short answer

To find the center of mass of a thin plate with unit density, which occupies the region between the parabola \(y = ax^2\) and the horizontal line \(y = b\), we perform the following steps: 1. Determine the mass of the region by finding the area of the region between the parabola and the horizontal line. The area A is given by: $$A = \int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} (b-ax^2)dx$$ 2. Find the x-coordinate of the center of mass. Due to the symmetry of the problem, the center of mass will lie on the y-axis, so the x-coordinate of the center of mass is 0. 3. Find the y-coordinate of the center of mass using the formula: $$\overline{y} = \frac{1}{A}\int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} \frac{1}{2}(b+ax^2)(b-ax^2)dx$$ Evaluating the integral and simplifying, the y-coordinate of the center of mass is found to be \(\frac{3b}{5}\). 4. The center of mass is at the point \(\left(0, \frac{3b}{5}\right)\) and it does not depend on the value of a.

Step-by-step solution

Determine the mass of the region

Since the region has unit density, we can calculate the mass by finding the area of the region between the parabola and the horizontal line. The area of this region can be determined using integration. We first need to find the x-intercepts of the parabola, where it intersects the horizontal line: $$ax^2 = b \Rightarrow x = \pm\sqrt{\frac{b}{a}}$$ The area A can be found by integrating over the region and considering that y=ax²: $$A = \int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} (b-ax^2)dx$$

Find the x-coordinate of the center of mass

The x-coordinate of the center of mass can be found using the formula: $$\overline{x} = \frac{1}{A}\int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} x(b-ax^2)dx$$ However, since the region is symmetric with respect to the y-axis, the center of mass will lie on the y-axis and hence \(\overline{x}\) = 0.

Find the y-coordinate of the center of mass

The y-coordinate of the center of mass can be found using the formula: $$\overline{y} = \frac{1}{A}\int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} \frac{1}{2}(b+ax^2)(b-ax^2)dx$$ Now we can solve the integral and the expression for the y-coordinate of the center of mass: $$\frac{1}{A}\int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} \frac{1}{2}(b^2 - a^2x^4)dx$$

Evaluate the integrals and find the center of mass

Evaluate the area integral in step 1: $$A = \int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} (b-ax^2)dx = 2\int_0^{\sqrt{\frac{b}{a}}} (b-ax^2)dx = 2 \left[bx - \frac{1}{3}ax^3\right]_0^{\sqrt{\frac{b}{a}}} = \frac{4}{3}b\sqrt{\frac{b}{a}}$$ Now we can evaluate the y-coordinate of the center of mass with the integral from step 3: $$\overline{y} = \frac{1}{\frac{4}{3}b\sqrt{\frac{b}{a}}}\int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} \frac{1}{2}(b^2 - a^2x^4)dx$$ $$= \frac{3}{4b\sqrt{\frac{b}{a}}}(b^2x - \frac{1}{5}a^2x^5)\Big|_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}}$$ $$= \frac{3}{4b\sqrt{\frac{b}{a}}}\left[\left(b^2\sqrt{\frac{b}{a}} - \frac{1}{5}a^2\left(\frac{b}{a}\right)^{\frac{5}{2}}\right) - \left(-b^2\sqrt{\frac{b}{a}} + \frac{1}{5}a^2\left(\frac{b}{a}\right)^{\frac{5}{2}}\right)\right]$$ $$= \frac{3}{4b\sqrt{\frac{b}{a}}}\left[2b^2\sqrt{\frac{b}{a}}\right] = \frac{3b}{5}$$ We have now found the x and y coordinates of the center of mass are \(\left(0, \frac{3b}{5}\right)\) and it does not depend on a.

Key concepts

Integration in Calculus

Integration in calculus is like finding the area under a curve. It's a core concept that's crucial for understanding how to calculate things like area, volume, and the center of mass. In this exercise, integration helps us determine the area between the parabola and the horizontal line. This area is essential because it represents the mass of the plate when it has unit density.

If you picture a function, it might look like curvy hills on a graph. To find out the size of this hill, we use integration to "sum up" the tiny parts under the curve. This process involves setting lower and upper limits, which are often the points where the curve touches or crosses the x-axis. In the example exercise, the limits are \(-\sqrt{\frac{b}{a}}\) and \(+\sqrt{\frac{b}{a}}\), showing where the parabola intersects the line.

Integration lets us handle more complex shapes beyond simple rectangles or triangles, making it a versatile tool in calculus. It's like a bridge to connect the algebra world with geometric and real-world visual problems.

Parabola

The parabola is a common, elegant shape in mathematics that can be perfectly described by an equation like \(y = ax^2\). Its classic U-shape is defined by its quadratic equation, and it shows up in various areas from physics to architecture.

In the scenario we're looking at, the parabola forms one boundary of the region whose mass we want to find. Understanding this shape is important because it helps define the limits for our integrals. The parabola symmetrically rests on the y-axis, making it straightforward to recognize its x-intercepts when solving calculus problems. These intercepts are key to identifying where the parabola meets the line \(y = b\), which gives us the region's boundaries.

Moreover, the parabola has unique properties such as a focus and directrix, and it's known for reflecting points symmetrically across its axis of symmetry. These properties are elegantly exploited in various applications when analyzing motions or designing structures.

Symmetry in Calculus

Symmetry in calculus is a powerful concept that simplifies complex problems. In this exercise, symmetry plays a pivotal role in determining the center of mass.

When we have symmetry, calculations tend to become more intuitive. For instance, in the exercise, the parabolic region is symmetrical about the y-axis. This symmetry means that the x-coordinate of the center of mass is 0. Why? Because each point on one side of the y-axis has an identical point on the opposite side, balancing out any differences.

Using symmetry efficiently enables us to focus only on the essential parts of the problem: the y-coordinate. This deep symmetry not only implies balance but also reduces computational effort, making calculations quicker and insights clearer. Recognizing and applying symmetry in calculus thus transforms challenging integrals into more manageable tasks.