Problem 12 - Thomas Calculus

Question

The integrals and sums of integrals in Exercises 9–14 give the areas of regions in the xy-plane. Sketch each region, label each bounding curve with its equation, and give the coordinates of the points where the curves intersect. Then find the area of the region. $$ \int_{-1}^{2} \int_{y^{2}}^{y+2} d x d y $$

Step-by-Step Solution

Answer

The area of the region is \( \frac{14}{3} \).

Step 1: Understand the Limits of Integration

The integral is given as \( \int_{-1}^{2} \int_{y^{2}}^{y+2} dx \, dy \), which means we first integrate with respect to \( x \) from \( x = y^2 \) to \( x = y + 2 \), and then with respect to \( y \) from \( y = -1 \) to \( y = 2 \). This setup suggests the bounding region is between these curves as viewed under the integration bounds.

Step 2: Sketch the Region

- The curve \( x = y^2 \) is a parabola opening to the right.- The line \( x = y + 2 \) is a straight line with a slope of 1 passing through the point \( (2,0) \).To sketch, draw both curves and note the region lies between them on the \( x \)-axis from \( y = -1 \) to \( y = 2 \).

Step 3: Find Intersection Points

To find where the curves intersect, set \( y^2 = y + 2 \). Solving the equation \( y^2 - y - 2 = 0 \), factor it into \( (y-2)(y+1) = 0 \), giving roots \( y = 2 \) and \( y = -1 \). These roots are the endpoints of the region.

Step 4: Set Up the Integral for Area

Since the integration limits are correct for finding the area between \( x = y^2 \) and \( x = y + 2 \), the given integral correctly describes the area of the region between these curves from \( y = -1 \) to \( y = 2 \).

Step 5: Calculate the Integral

Perform the integration:Evaluate the inner integral:\[ \int_{y^2}^{y+2} dx = (y+2) - (y^2) = -y^2 + y + 2 \]Evaluate the outer integral:\[ \int_{-1}^{2} (-y^2 + y + 2) \, dy \]Calculate each term separately:\[ \int_{-1}^{2} -y^2 \, dy = \left[ -\frac{y^3}{3} \right]_{-1}^{2} = \left( -\frac{8}{3} - \left(-\frac{-1}{3}\right)\right) = -\frac{8}{3} + \frac{1}{3} = -\frac{7}{3} \]\[ \int_{-1}^{2} y \, dy = \left[ \frac{y^2}{2} \right]_{-1}^{2} = 2 - \frac{1}{2} = \frac{4}{3} \]\[ \int_{-1}^{2} 2 \, dy = 2 \cdot 3 = 6 \]Sum the results: \[-\frac{7}{3} + \frac{9}{6} + 6 = \frac{-7 + 3 + 18}{3} = \frac{14}{3} \]

Step 6: Conclusion and Final Area

The area of the region bounded by the curves \( x = y^2 \) and \( x = y + 2 \) between \( y = -1 \) and \( y = 2 \) is calculated to be \( \frac{14}{3} \).

Key Concepts

Area of Regions

Double integrals are powerful tools when calculating the area of regions in the xy-plane. Here, the main task is to assess the space between curves. For example, if you're looking at an integration \[ \int_{-1}^{2} \int_{y^{2}}^{y+2} dx \, dy \], you're determining the area between the parabolic curve \( x = y^2 \) and the linear curve \( x = y + 2 \), from \( y = -1 \) to \( y = 2 \). Using this setup, we are able to calculate the total area enclosed by these curves. This method allows us to break down complex regions into manageable parts, evaluated within the given boundaries.

Intersection Points

Finding the intersection points before evaluating a double integral is crucial in determining the limits of integration. Here, we set the equations \( y^2 = y + 2 \) equal to each other to find the intersection points. Solving this gives us the points \( y = 2 \) and \( y = -1 \). These points are essential because they define where the curves meet; in other words, they determine the vertical limits for our integral, guiding both the setup and evaluation of our problem.

Parabolic and Linear Curves

Understanding the nature of the curves involved is key to sketching the region and setting up the integration properly. - The curve \( x = y^2 \) represents a parabola that opens to the right. Its vertex is at the origin and its path curves upwards and downwards symmetrically from there.- The line \( x = y + 2 \) is a straightforward linear equation with a slope of 1, crossing the y-axis at the point (0,2).Together, these curves form the boundaries of the region whose area we calculate using double integrals. Understanding these will help visualize the problem and ensure the integral is set up accurately.

Integration Limits

Integration limits are essential in defining the scope of your calculation. For the double integral equation \( \int_{-1}^{2} \int_{y^{2}}^{y+2} dx \, dy \), the limits are determined by 1. The outer integration limits corresponding to \( y \), which run from \( -1 \) to \( 2 \), 2. The inner limits corresponding to \( x \), which go from \( y^2 \) to \( y + 2 \).These specify the bounds over which the integration operates. The limits must accurately reflect the boundaries of the enclosed region, ensuring the correct area is calculated by the integral. Reviewing and understanding these limits is crucial for accurate computation.