Problem 42 - Calculus Volume 1

Question

For the following exercises, find the exact area of the region bounded by the given equations if possible. If you are unable to determine the intersection points analytically, use a calculator to approximate the intersection points with three decimal places and determine the approximate area of the region. $$ [T]x^{2}=y^{3} \text { and } x=3 y $$

Step-by-Step Solution

Answer

The approximate area of the bounded region is 461.7 square units.

Step 1: Understand the Problem

We are given two curves defined by the equations \( x^2 = y^3 \) and \( x = 3y \). We need to find the region enclosed by these two curves and calculate its area.

Step 2: Find the Intersection Points

To find the intersection points of the two curves, set \( x^2 = y^3 \) equal to \( x = 3y \). First, express \( x \) from the second equation in terms of \( y \): \( x = 3y \). Substitute this into the first equation to get \( (3y)^2 = y^3 \). This simplifies to \( 9y^2 = y^3 \), or \( y^3 - 9y^2 = 0 \). Factor to find \( y(y^2 - 9y) = 0 \), giving solutions \( y = 0 \) or \( y = 9 \). Substitute these back into \( x = 3y \) to find the intersection points: \((0, 0)\) and \((27, 9)\).

Step 3: Set Up the Integral for the Area

The area between the curves can be found by integrating the difference between the top function and the bottom function from the intersection points. In terms of \( y \), the function \( x = y^{3/2} \) from \( x^2 = y^3 \) (for \( y \geq 0 \)) and \( x = 3y \) represent the curves. The limits of integration are from \( y = 0 \) to \( y = 9 \). The area \( A \) is given by the integral \[ A = \int_{0}^{9} (3y - y^{3/2}) \, dy \].

Step 4: Calculate the Integral

To find the integral, compute: \[ \int (3y - y^{3/2}) \, dy = \left[ \frac{3y^2}{2} - \frac{2}{5}y^{5/2} \right]_{0}^{9} \]. Evaluate this expression at the bounds: \[ \left( \frac{3(9)^2}{2} - \frac{2}{5}(9)^{5/2} \right) - \left( \frac{3(0)^2}{2} - \frac{2}{5}(0)^{5/2} \right) = \frac{243}{2} - \frac{2916}{5} \].

Step 5: Simplify and Interpret the Result

Simplify the expression \( \frac{243}{2} - \frac{2916}{5} \) to find the exact area. Converting each fraction to a common denominator (10), the calculation becomes: \[ \frac{243}{2} = \frac{1215}{10}, \quad \frac{2916}{5} = \frac{5832}{10} \]. So, the integral evaluates to \[ \left( \frac{1215}{10} - \frac{5832}{10} \right) = -\frac{4617}{10} \]. Since area cannot be negative, take the absolute value: \[ \frac{4617}{10} = 461.7 \]. This is the approximate area of the bounded region.

Key Concepts

Intersection Points

Finding intersection points of two curves is a crucial step in determining the area they enclose. It tells us where the curves meet and therefore bounds the region of interest. In our problem, the curves given are \( x^2 = y^3 \) and \( x = 3y \). To find their intersection points, we equate the two equations for \( x \) values.

• Start with \( 9y^2 = y^3 \) by substituting \( x = 3y \) into the first equation.
• Solves to \( y(y - 9) = 0 \) after factoring out \( y^2 \).
• The solutions \( y = 0 \) and \( y = 9 \) give us the points \((0,0)\) and \((27,9)\) when substituted back into \( x = 3y \).

These points mark where the curves cross and help define the limits for our area calculation. Knowing intersection points ensures that we integrate over the correct interval. Without them pinpointed correctly, the resulting area might be inaccurate.

Area Under Curve

The area under a curve, particularly between two curves, represents the total measure of the space enclosed. In calculus, this is often solved by finding the definite integral of a function. Considering our example:

• First, determine the functions themselves: \( x = y^{3/2} \) representing \( x^2 = y^3 \) and \( x = 3y \).
• Then, set up an equation for the area \( A \) between these two functions from their intersection points \((0, 0)\) to \((27, 9)\).

To calculate the area, integrate the top function minus the bottom function, with respect to \( y \), from the lower intersection point to the upper:\[A = \int_{0}^{9} (3y - y^{3/2}) \, dy\]The integral of this difference gives the area between the curves from point to point. This type of calculation allows for an exact measure of the space bounded by such curves.

Integration Techniques

Integration techniques allow us to calculate the area between or under curves by evaluating definite integrals. There are various techniques that one might use in this process, depending on the functions involved.

• Understand first the concept of integration: It's essentially summing an infinite number of infinitesimally small quantities.
• For our particular problem, integrate the difference \(3y - y^{3/2}\).

Evaluate the integral step-by-step to find:\[\int (3y - y^{3/2}) \, dy = \left[ \frac{3y^2}{2} - \frac{2}{5}y^{5/2} \right]_{0}^{9}\]The result involves:

• Plugging in the boundaries \( y = 0 \) and \( y = 9 \) into the antiderivative to find their respective areas.
• Subtracting these areas to find the net result over the interval.
• Taking the absolute value, as area cannot be negative.

Utilizing integration techniques and applying the fundamental theorem of calculus turns abstract problems into tangible solutions, illustrating the power and usefulness of integrals in determining areas.