Problem 16 - Thomas Calculus

Question

The integrals and sums of integrals in Exercises \(13 - 18\) give the areas of regions in the \(x y\) -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

Area = \(\frac{9}{2}\) square units.

Step 1: Understanding the Domain

The given integral \(\int _{ - 1 } ^{ 2 } \int _{ y ^{ 2 } } ^{ y + 2 } dx \, dy\) describes a region in the \(xy\)-plane. We interpret this as integrating over \(y\) from \(-1\) to \(2\), and for each \(y\), the \(x\)-values range from \(y^2\) to \(y + 2\). Our task is to sketch and identify this region.

Step 2: Sketching the Curves

The bounding curves are \(x = y^2\) and \(x = y + 2\). These represent a parabola opening to the right and a straight line with a slope of 1. We will sketch these curves in the \(xy\)-plane. The curve \(x = y^2\) is symmetric about the x-axis, while \(x = y + 2\) is a diagonal line.

Step 3: Finding Intersection Points

To find intersection points, set \(y^2 = y + 2\). This simplifies to \(y^2 - y - 2 = 0\). Factoring gives \((y - 2)(y + 1) = 0\). Thus, \(y = 2\) and \(y = -1\). Substituting back, the intersections are \((4, 2)\) and \((1, -1)\).

Step 4: Describing the Region

With the intersections identified, the region of interest is defined by \(y\) going from \(-1\) to \(2\) and \(x\) ranging between \(x = y^2\) (left boundary) and \(x = y+2\) (right boundary). It is the vertical strip between the parabola and line over this \(y\) range.

Step 5: Calculating the Area

The integral \(\int _{ -1 } ^{ 2 } \int _{ y^2 } ^{ y + 2 } dx \, dy\) computes the area of the region. We calculate the inner integral first: \(\int _{ y^2 } ^{ y+2 } 1 \, dx = x \biggr|_{y^2}^{y+2} = (y+2) - (y^2)\). This simplifies the double integral to: \(\int _{-1}^{2} (y+2 - y^2) \, dy \). Now compute: \[ \int _{-1}^{2} (y+2 - y^2) \, dy = \int _{-1}^{2} y \, dy + \int _{-1}^{2} 2 \, dy - \int _{-1}^{2} y^2 \, dy \].

Step 6: Evaluating Individual Integrals

Compute each part: 1. \(\int _{-1}^{2} y \, dy = \frac{1}{2}y^2 \biggr|_{-1}^{2} = 2 - \frac{1}{2} = \frac{3}{2}\). 2. \(\int_{-1}^{2} 2 \, dy = 2y \biggr|_{-1}^{2} = 6\).3. \(\int _{-1}^{2} y^2 \, dy = \frac{1}{3}y^3 \biggr|_{-1}^{2} = \frac{8}{3} - \left(-\frac{1}{3}\right) = \frac{9}{3} = 3\).Summing these results: \(\frac{3}{2} + 6 - 3 = \frac{3}{2} + 3 = \frac{9}{2}\).

Step 7: Final Result

Add the results from step 6: The total area is \(\frac{9}{2}\) square units.

Key Concepts

Region in the XY-Plane

When dealing with double integrals, one key concept is understanding the region over which you're integrating. This is essential because it determines the limits of integration and helps us visualize what we are calculating. In this problem, we look at \( \int _ { - 1 } ^ { 2 } \int _ { y ^ { 2 } } ^ { y + 2 } d x d y \), which defines a region in the \( xy \)-plane. This means our interest lies in the two-dimensional space bounded by these limits. The integral describes iterating over \( y \) from \(-1 \) to \( 2 \), defining the vertical span, and for each specific value of \( y \), \( x \) values range from \( y^2 \) to \( y + 2 \). Knowing this, you can visualize the domain as a vertical strip that has a pre-defined bottom edge, top edge, left boundary, and right boundary.

Intersection Points

In this exercise, finding where the two curves intersect is crucial. It allows us to determine the exact region enclosed by these curves. We look at the equations of the bounding curves:

• \( x = y^2 \), a parabola opening to the right,
• \( x = y + 2 \), a straight line.

To find their points of intersection, we set the equations equal, \( y^2 = y + 2 \), and solve for \( y \). Simplifying gives a quadratic \( y^2 - y - 2 = 0 \). Upon factoring, we get \( (y - 2)(y + 1) = 0 \), leading to solutions \( y = 2 \) and \( y = -1 \). These values substitute back into either equation to find corresponding \( x \)-values, giving points \((4, 2)\) and \((1, -1)\) as the intersections. These points help us neatly define the region of interest.

Area Calculation

Calculating the area of the region enclosed by the curves is the goal of our integral. Once we identify the region in the \( xy \)-plane and the bounds for \( y \) and \( x \), we can set up our double integral to find the area. Our task here includes:1. Evaluating the iterated integral in steps.2. First, perform the inner integral: \( \int _{ y^2 } ^{ y+2 } 1 \, dx \), which simplifies to \((y+2 - y^2)\).3. Then, integrate this result over \( y \) from \(-1\) to \( 2 \): \[ \int _{-1}^{2} (y+2 - y^2) \, dy \].Breaking it down, integrate each part separately to get the sum \( \frac{9}{2} \) square units as the total area. It involves some basic integration rules to compute these efficiently, ensuring every student can follow along and replicate at this point.

Bounding Curves

Bounding curves are essential in analyzing and solving integration problems that deal with regions in the \( xy \)-plane. For this exercise, our bounding curves are \( x = y^2 \) and \( x = y + 2 \). Understanding how each boundary functions helps to accurately describe the area we wish to calculate.

• The curve \( x = y^2 \) forms a parabola opening to the right and acts as the left boundary in our region.
• The curve \( x = y + 2 \) is a straight line with a positive linear slope, serving as the right boundary.

Between \( y = -1 \) and \( y = 2 \), the region of interest stretches horizontally between these curves. By setting up and solving the problems with these bounding curves in mind, students can see their role in determining the region over which we perform the integration.