Problem 16 - University Calculus: Early Transcendentals

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

The area is \(\frac{9}{2}\) or 4.5 square units.

Step 1: Understand the Integral Setup

The given double integral \( \int_{-1}^{2} \int_{y^{2}}^{y+2} dx \, dy \) represents the area of a region in the \(xy\)-plane. The bounds of integration for \(x\) are from \(y^2\) to \(y+2\), and \(y\) ranges from \(-1\) to \(2\). This indicates that the region is bounded horizontally between \(x = y^2\) and \(x = y + 2\), and vertically from \(y = -1\) to \(y = 2\).

Step 2: Sketch the Region

Draw the curves \(x = y^2\) and \(x = y + 2\). Plot them on the \(xy\)-plane. Notice that the parabola \(x = y^2\) opens to the right, and the line \(x = y + 2\) is a straight line with a slope of 1. The vertical limits \(y=-1\) and \(y=2\) help define the region of interest.

Step 3: Find Intersection Points

To find where these curves intersect, solve for \(y\) in the equation \(y^2 = y + 2\). Rearrange to get \(y^2 - y - 2 = 0\). Factoring gives \((y - 2)(y + 1) = 0\), leading to the solutions \(y = 2\) and \(y = -1\). The corresponding \(x\)-values are found by substituting back into either equation: for \(y = 2\), \(x = 4\) and for \(y = -1\), \(x = 1\). The intersection points are \((4, 2)\) and \((1, -1)\).

Step 4: Set Up and Evaluate the Integrals

For the double integral, evaluate the inner integral first: \[ \int_{y^2}^{y+2} 1 \, dx = (y + 2) - (y^2). \]This simplifies to \(-y^2 + y + 2\). Next, integrate this expression with respect to \(y\) from \(-1\) to \(2\):\[ \int_{-1}^{2} (-y^2 + y + 2) \, dy.\]

Step 5: Calculate the Result

Calculate the integral \[ \int (-y^2 + y + 2) \, dy = \left[ -\frac{y^3}{3} + \frac{y^2}{2} + 2y \right]_{-1}^{2}. \]Substitute the bounds: \( \left(-\frac{8}{3} + 2 + 4) - \left(\frac{1}{3} - \frac{1}{2} - 2\right) = \frac{9}{2}.\)

Step 6: Conclude the Area

The final result, \(\frac{9}{2}\), represents the area of the region bounded by the given curves \(x = y^2\), \(x = y + 2\), and the horizontal lines \(y = -1\) and \(y = 2\).

Key Concepts

Area in xy-plane

When dealing with double integrals, a common application is to find the area of a region in the xy-plane. Double integrals allow us to sum infinitesimal elements of area over a specified region. The region itself is defined by the boundaries of the integral. Thus, understanding how these integrals translate to areas is crucial.

To compute the area, we visualize the region enclosed by the given boundaries. The area is represented by the double integral's setup, which specifies the limits for the integration variables. In this case, the double integral \( \int_{-1}^{2} \int_{y^{2}}^{y+2} dx \, dy \) represents an area bounded by the curves and lines defined by the integration limits. This approach helps us turn conceptual boundary definitions into precise area calculations.

Curve Intersection

Understanding where curves intersect is vital because it helps define the limits of integration for certain problems. Here, we have two curves: \( x = y^2 \) and \( x = y + 2 \).

To find the points where these curves intersect, we solve for \( y \) in the equation \( y^2 = y + 2 \). By rearranging terms, we arrive at the quadratic equation \( y^2 - y - 2 = 0 \). Factoring gives us \( (y - 2)(y + 1) = 0 \), leading to the solutions \( y = 2 \) and \( y = -1 \). Substituting these values back into either equation gives us the \( x \)-coordinates, resulting in the points \( (4, 2) \) and \( (1, -1) \). These intersection points help determine the boundaries for the integration.

Integration Limits

Integration limits define the range over which the function is integrated. These limits can be functions themselves, as seen in the problem \( \int_{-1}^{2} \int_{y^{2}}^{y+2} dx \, dy \).

The inner integral limits \( y^2 \) to \( y+2 \) describe how \( x \) varies with \( y \). This means for every given \( y \) within its limits, \( x \) ranges between \( y^2 \) to \( y+2 \).

The outer integral limits of \( y \) from \(-1\) to \(2\) describe how far vertically the calculated area extends. Careful attention to these limits ensures we capture the full extent of the region in question. Integration limits are fundamental in ensuring the area covered is accurate and corresponds to the realistic scenario described by the problem.

Bounding Equations

Bounding equations are equations that define the edges of the region we're interested in. They form the perimeter of the area for integration.

In our exercise, the bounding equations are \( x = y^2 \) and \( x = y + 2 \). These equations help create a picture of the space by representing the boundaries on the xy-plane.

Graphing these lines can clarify the concept: \( x = y^2 \) is a right-opening parabola, while \( x = y + 2 \) is a straight line with a slope of 1. These graphs intersect at two points, which are critical to determining the limits and hence, the area. The multiple intersections combined with vertical boundaries \( y = -1 \) and \( y = 2 \) close the region, making these equations vital to formulating, setting, and solving the double integral for the area calculation.