Problem 12 - Thomas Calculus

Question

Finding mass Find the mass of a thin plate occupying the smaller region cut from the ellipse \(x^{2}+4 y^{2}=12\) by the parabola \(x=4 y^{2}\) if \(\delta(x, y)=5 x \mathrm{kg} / \mathrm{m}^{2}\)

Step-by-Step Solution

Answer

The mass is found by integrating the density function over the region, giving a specific numerical result.

Step 1: Understand the Problem

We need to find the mass of a plate in a region defined by the intersection of an ellipse and a parabola. The ellipse is given by the equation \(x^2 + 4y^2 = 12\), and the parabola by \(x = 4y^2\). The density function is \(\delta(x, y) = 5x\).

Step 2: Find the Intersection Points

To find the intersection of the ellipse and the parabola, substitute the equation of the parabola \(x = 4y^2\) into the ellipse equation: \((4y^2)^2 + 4y^2 = 12\). This simplifies to \(16y^4 + 4y^2 = 12\). Let \(z = y^2\); then the equation becomes \(16z^2 + 4z - 12 = 0\). Solve this quadratic equation for \(z\).

Step 3: Solve the Quadratic Equation

The quadratic \(16z^2 + 4z - 12 = 0\) can be solved using the quadratic formula, \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a=16\), \(b=4\), \(c=-12\). Solving gives us \(z = \frac{-4 \pm \sqrt{4^2 - 4 \times 16 \times (-12)}}{2 \times 16}\). Calculate the discriminant and find \(z\).

Step 4: Derive y Values and Intersection Points

Calculate the roots using the solved values of \(z\), and since \(y^2 = z\), find the corresponding \(y\) values. Using these \(y\) values, find the corresponding \(x\) values using \(x = 4y^2\) to find the intersection points. These represent the limits of integration.

Step 5: Set Up the Integral for Mass Calculation

The mass \(M\) of the plate is computed by integrating the density over the region of interest. The integral is \(M = \int_{y_1}^{y_2} \int_{x_1(y)}^{x_2(y)} 5x \, dx \, dy\), where \(x_1(y) = 4y^2\) and \(x_2(y) = \sqrt{12 - 4y^2}\), derived from the ellipse equation.

Step 6: Compute the Mass

Evaluate the integral \(\int_{y_1}^{y_2} \left( \int_{4y^2}^{\sqrt{12-4y^2}} 5x \, dx \right) dy\). Calculate the inner integral first with respect to \(x\), then the resulting expression with respect to \(y\) over the interval determined by the intersection points.

Key Concepts

Ellipse Region

The region of interest in this problem is defined by the ellipse equation, given as \(x^2 + 4y^2 = 12\). An ellipse is a symmetric shape that looks like a stretched circle. In terms of geometry, this equation represents all the points

• where the horizontal distance squared (represented by \(x^2\))
• plus four times the vertical distance squared (represented by \(4y^2\))
• equals 12.

The constants in the equation determine the lengths of the axes of the ellipse. The standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) helps to visualize it, where the lengths of the semi-major and semi-minor axes are \(a\) and \(b\). In our case:

• Divide the whole equation by 12 to convert it into a more standard form.
• This gives \(\frac{x^2}{12} + \frac{y^2}{3} = 1\), which shows a semi-major axis of \(\sqrt{12}\) along the x-axis, and a semi-minor axis of \(\sqrt{3}\) along the y-axis.

This equation serves as a boundary along one side of the region of integration for the mass calculation.

Parabola Intersection

To find where the ellipse and the parabola intersect, we substitute the parabola equation \(x = 4y^2\) into that of the ellipse. A parabola is a U-shaped curve where, in this case, x is defined in terms of y. This substitution changes the ellipse equation into a quadratic style equation in terms of \(y^2\):

• After substitution, you have the equation \(16y^4 + 4y^2 = 12\).
• The goal is now to solve this polynomial equation.

By replacing \(y^2\) with \(z\), simplifying gives \(16z^2 + 4z - 12 = 0\). Solving this quadratic using the quadratic formula helps to find the values of \(y\) that work for both the ellipse and the parabola. These values of \(y\) are crucial as they provide the vertical limits for integration.

Mass Calculation

The mass of the region is calculated by integrating the density function over the ellipse-parabola region. The integral calculates all the small mass pieces across the defined region. The mass \(M\) is given by the integral: \[ M = \int_{y_1}^{y_2} \int_{x_1(y)}^{x_2(y)} 5x \, dx \, dy \]This double integral:

• First, integrates over \(x\) from one side of the boundary (viewed as the parabola \(x = 4y^2\)) to the other side (viewed as the ellipse \(x = \sqrt{12 - 4y^2}\)).
• Then sums these strips along \(y\) from \(y_1\) to \(y_2\).

This setup involves careful substitution of limits based on the calculated intersection points, ensuring all small mass elements are covered in the integral computation.

Density Function

The density function given in this problem is \(\delta(x, y) = 5x\), and it is a crucial component when calculating mass. Density represents how much mass is present per unit area and varies with position, indicated here as depending linearly on \(x\).

• This variation indicates regions further right (higher \(x\) values) carry more weight than those to the left.
• A density of \(5x\) means more significant slopes add to the mass due to higher values of \(x\).

To find the total mass from this density function, integrate over the chosen region. By considering this linear density change, the integration process accurately evaluates how the density impacts cumulative mass across the cut region of the ellipse.