Problem 23 - Technical Mathematics with Calculus

Question

Find the area between the parabolas \(y^{2}=2 x\) and \(x^{2}=2 y\)

Step-by-Step Solution

Answer

Evaluate the definite integral of the difference of the two functions from 0 to 2, which results in the area between the parabolas being \(\frac{4}{3}\).

Step 1: Graph both equations

Begin by sketching the graphs of the parabolas. The graph of the parabola defined by the equation \(y^{2} = 2x\) opens to the right, whereas the graph of the parabola defined by \(x^{2} = 2y\) opens upwards. Identify the points of intersection by setting the equations equal to each other, which will give you the bounds for the area.

Step 2: Find the points of intersection

To find the points of intersection, equate \(y^2 = 2x\) to \(x^2 = 2y\). By substitution, we get \(x^2 = 2y = 2(\sqrt{2x})\). By simplifying, you'll get an expression \(x = 0\) or \(x = 2\), which correspond to the points of intersection, (0,0) and (2,2).

Step 3: Set up the integral to find the area

We will evaluate the area by integrating the difference between the two functions from the leftmost to the rightmost point of intersection. The integral setup would be \(A = \int_{0}^{2}(\sqrt{x^2/2}-\sqrt{2x})dx\), where \(\sqrt{x^2/2}\) is the top function (from the equation \(x^2 = 2y\)) and \(\sqrt{2x}\) is the bottom function (from the equation \(y^2 = 2x\)).

Step 4: Evaluate the integral

Evaluating the integral \(A = \int_{0}^{2}(\sqrt{x^2/2}-\sqrt{2x})dx\) requires calculating the antiderivative and evaluating from 0 to 2. By computing the integral, we find the function for area A(x), and then find A(2) - A(0) to get the area between the curves.

Step 5: Simplify the calculation

The antiderivative can be simplified using u-substitution or recognizing the derivative forms in the integrand, obtaining \(A(x) = \frac{1}{2}(x\sqrt{x^2/2} - x)\), from which we can plug in the limits of integration.

Step 6: Find the definite integral values

After finding the antiderivative, plug in the upper limit x = 2 and subtract the value when plugging in the lower limit x = 0. The result will give you the area between the parabolas.

Key Concepts

Graphing Parabolas

Understanding how to graph parabolas is vital when solving problems involving the area between curves.

Parabolas are U-shaped graphs that can open upwards, downwards, to the left, or to the right, depending on their equation. If the equation is in the form of \( y^2 = 4ax \) or \( x^2 = 4ay \), the parabola opens to the right or upwards respectively if \( a \) is positive, and to the left or downwards if \( a \) is negative.

Key Tips for Graphing Parabolas:

• Identify the vertex: This is the turning point of the parabola, found at the origin in its standard equation.
• Determine the direction: Look at the coefficient of \( x^2 \) or \( y^2 \) to see which way the parabola opens.
• Find the focal point and directrix as additional points that define the shape.

When you graph the given parabolas \( y^2 = 2x \) and \( x^2 = 2y \) on the same axes, they reveal the area trapped between them, which is precisely what we are looking to calculate.

Points of Intersection

The points of intersection of two curves are where their graphs meet or cross over.

To find these points algebraically, set the equations equal to each other and solve for the variable. In this case, we have the equations \( y^2 = 2x \) and \( x^2 = 2y \) which, when equated, allow us to find the coordinates where the parabolas intersect.

Steps to Determine the Points of Intersection:

• Set up the equations to be equal: \( y^2 = x^2/2 \).
• Solve for \( x \) or \( y \) by isolating the variable and taking square roots if necessary.
• Check the solutions in both original equations to ensure they satisfy both, eliminating any extraneous solutions.

This process results in the points (0,0) and (2,2) for our parabolas, establishing the bounds for the area we want to find.

Definite Integrals

Definite integrals are fundamental tools in calculus used to calculate areas, volumes, and other concepts linked with accumulation.

A definite integral is represented as \( \int_{a}^{b} f(x) dx \) and it computes the net area under the curve \( f(x) \) from \( x=a \) to \( x=b \) on a graph. When the function is above the x-axis, this area is positive, and when it’s below, it's negative.

Application in Finding Areas:

• To find the area between two curves, set up a definite integral with the upper bound and lower bound as the points of intersection.
• The integrand is the difference between the top and bottom functions.
• Evaluating the definite integral gives the area between the curves.

In our case, the integral \( \int_{0}^{2} (\sqrt{x^2/2}-\sqrt{2x}) dx \) was used to find the area between the parabolas, with the limits from the points of intersection.

U-Substitution Method

The u-substitution method is an algebraic technique used to simplify the process of finding antiderivatives, which is necessary when evaluating definite integrals.

This method involves choosing a piece of the integral as ‘u’ to make the integral simpler and resemble a form we can integrate. Consider it a way of transforming a complicated integral into a basic one.

Key Steps in U-Substitution:

• Select \( u \) to be a function inside the integral that appears multiple times or makes deriving easier.
• Differentiate \( u \) to find \( du \) and solve for \( dx \) or vice versa.
• Substitute all instances of \( x \) in the integral with \( u \) and \( dx \) with \( du \) to get an integral in ‘u’.
• Integrate with respect to \( u \), and then substitute back to return to the original variable.

For our problem, u-substitution can simplify the integral to find the antiderivative, paving the way to easily computing the area between the parabolas.