Problem 7 - Thomas Calculus

Question

In Exercises 1–8, sketch the region bounded by the given lines and curves. Then express the region’s area as an iterated double integral and evaluate the integral. The parabolas \(x=y^{2}\) and \(x=2 y-y^{2}\)

Step-by-Step Solution

Answer

The area of the bounded region is \(\frac{1}{3}\).

Step 1: Find the Points of Intersection

To find where the parabolas intersect each other, we set the equations equal: \[ y^2 = 2y - y^2 \]Simplifying, we combine like terms: \[ y^2 + y^2 - 2y = 0 \]This simplifies to: \[ 2y^2 - 2y = 0 \]Factoring out a 2y gives:\[ 2y(y - 1) = 0 \]Thus, the solutions are \( y = 0 \) and \( y = 1 \). These provide the intersection points \((0, 0)\) and \((1, 1)\) when plugged back into either original equation.

Step 2: Sketch the Region

We sketch the parabolas and identify the region bounded by them. The parabola \( x = y^2 \) opens rightward, and the parabola \( x = 2y - y^2 \) opens leftward.With intersection points at \((0,0)\) and \((1,1)\), the bounded region is between these curves in the segment \(0 \leq y \leq 1\).

Step 3: Set Up the Double Integral

The region is bounded horizontally by the curves \(x = y^2\) (left curve) and \(x = 2y - y^2\) (right curve). The vertical bounds are \(0\) and \(1\) (from the y-values of intersection points). The area \(A\) can be expressed as an iterated integral:\[A = \int_{0}^{1} \int_{y^2}^{2y-y^2} dx \, dy\]

Step 4: Evaluate the Inner Integral

The inner integral, with respect to \(x\), is:\[\int_{y^2}^{2y-y^2} dx = [x]_{y^2}^{2y-y^2} = (2y-y^2) - y^2\]Simplifying:\[2y - 2y^2\]

Step 5: Evaluate the Outer Integral

Substitute the result of the inner integral into the outer integral:\[\int_{0}^{1} (2y - 2y^2) \, dy\]Calculate this integral:\[\int_{0}^{1} (2y - 2y^2) \, dy = \left[y^2 - \frac{2}{3}y^3 \right]_{0}^{1}\]Evaluating from 0 to 1:\[\left(1 - \frac{2}{3}(1)^3\right) - \left(0 - \frac{2}{3}(0)^3\right) = 1 - \frac{2}{3} = \frac{1}{3}\]

Step 6: Conclusion

The area of the region bounded by the parabolas \(x = y^2\) and \(x = 2y - y^2\) is \(\frac{1}{3}\). This is derived by evaluating the double integral set up from the expressions for the boundaries.

Key Concepts

Intersection of Curves

When solving problems involving the intersection of curves, the main goal is to find the points where two or more curves meet. This is crucial because these points define the boundaries of the region we want to study or calculate. In this exercise, the curves given are two parabolas: \( x = y^2 \) and \( x = 2y - y^2 \). To find their intersection points, we equate the two expressions:

• \( y^2 = 2y - y^2 \)

This is a common technique where we solve for one variable in terms of another, in this case \( y \). Simplifying gives:

• \( 2y^2 - 2y = 0 \)
• Factoring gives us \( 2y(y - 1) = 0 \), so \( y = 0 \) or \( y = 1 \).

These solutions correspond to the ordered pairs \((0,0)\) and \((1,1)\) when plugged back into either parabola equation. These points of intersection help us establish the limits for the region bounded by the curves. These boundaries are foundational for setting up our iterated integral.

Area of Bounded Regions

The area of a region bounded by curves can be calculated using double integrals. Once we've determined the points of intersection, we identify the full area enclosed by the curves. Imagine shading the area between the two parabolas on a graph. In this scenario, the curves \(x = y^2\) and \(x = 2y - y^2\) form a region that is horizontally bounded.The left boundary is described by the curve \(x = y^2\), and the right boundary is \(x = 2y - y^2\). Vertically, the region is bounded between the intersection points' y-values, from \(y = 0\) to \(y = 1\). To express this area as an integral, we can set up an iterated integral:

• \[A = \int_{0}^{1} \int_{y^2}^{2y-y^2} dx \, dy\]

This integral will allow us to calculate the total area by integrating in the \(x\)-direction first between the curves and then in the \(y\)-direction over the specified bounds.

Integration Techniques

Evaluating double integrals involves integration techniques that apply to both inner and outer integrals. The process begins with the inner integral, which is set between the horizontal boundaries (between \(x = y^2\) and \(x = 2y-y^2\)) in this problem.

• The inner integral: \[ \int_{y^2}^{2y-y^2} dx \]
• This simplifies to \(2y - 2y^2\) after evaluating the difference between the upper and lower limits of \(x\).

After completing the inner integral, the next step is to substitute this simplified expression into the outer integral and evaluate it:

• \[ \int_{0}^{1} (2y - 2y^2) \, dy \]

Integrating with respect to \(y\) involves finding the antiderivative:

• Calculate \( y^2 - \frac{2}{3}y^3 \).

Finally, substituting the limits from 0 to 1 will give us the exact area, which, in this instance, is \( \frac{1}{3} \). Understanding how to approach these integrals and apply techniques correctly is crucial for simplifying these types of problems efficiently.