Problem 14 - 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. $$\int_{0}^{3} \int_{-x}^{\sqrt{(2-x)}} d y d x$$

Step-by-Step Solution

Answer

The area of the region is \( \frac{9}{2} + \frac{4\sqrt{2}}{3} \).

Step 1: Identify the bounding curves

The given double integral is \( \int_{0}^{3} \int_{-x}^{\sqrt{2-x}} dy \, dx \). This represents the region bounded by the curves \( y = -x \), \( y = \sqrt{2-x} \), and the vertical lines \( x = 0 \) and \( x = 3 \).

Step 2: Find the intersection points

Set \( -x = \sqrt{2-x} \) to find the intersection points. Square both sides to get \( x^2 = 2 - x \). Simplify to get \( x^2 + x - 2 = 0 \). Factor the quadratic to \( (x-1)(x+2) = 0 \). The solutions are \( x = 1 \) and \( x = -2 \), but since \( x \) ranges from 0 to 3, only \( x = 1 \) is valid. Solve for \( y \) to get \( y = -1 \) when \( x = 1 \). Thus, the intersection point is \((1, -1)\).

Step 3: Sketch the region

Draw the region in the xy-plane. Plot the curves \( y = -x \) and \( y = \sqrt{2-x} \) along with the vertical bounds \( x = 0 \) and \( x = 3 \). The region of interest is between these curves from \( x = 0 \) to \( x = 3 \).

Step 4: Compute the double integral to find the area

Evaluate the integral \( \int_{0}^{3} \int_{-x}^{\sqrt{2-x}} dy \, dx \) to find the area. Integrate with respect to \( y \) first, resulting in \( \int_{-x}^{\sqrt{2-x}} 1 \ dy = [y]_{-x}^{\sqrt{2-x}} = \sqrt{2-x} + x \). Then, evaluate the integral with respect to \( x \): \( \int_{0}^{3} (\sqrt{2-x} + x) \, dx \).

Step 5: Integrate with respect to x

Separate \( \int_{0}^{3} (\sqrt{2-x} + x) \, dx = \int_{0}^{3} \sqrt{2-x} \, dx + \int_{0}^{3} x \, dx \). Calculate \( \int \sqrt{2-x} \, dx \) using substitution: let \( u = 2-x \), then \( du = -dx \). Change the limits for \( u \) from 2 to -1, resulting in \( -\int_{2}^{-1} \sqrt{u} \, du \). The antiderivative of \( \sqrt{u} \) is \( \frac{2}{3}u^{3/2} \). Evaluate this from 2 to -1 and adjust for the negative to find the area.

Step 6: Complete the integration

Evaluating \( \int_{0}^{3} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{3} = \frac{9}{2} \). Compute \( -\left[ \frac{2}{3}u^{3/2} \right]_{2}^{0} = -\left(\frac{2}{3}(0) - \frac{2}{3}(2^{3/2})\right) = \frac{4\sqrt{2}}{3} \). Add \( \frac{4\sqrt{2}}{3} + \frac{9}{2} \) to find the total area.

Key Concepts

Bounding Curves

When working with double integrals, identifying the bounding curves is crucial. For this exercise, the double integral \( \int_{0}^{3} \int_{-x}^{\sqrt{2-x}} dy \, dx \) helps us find the area of a region in the xy-plane. The bounds consist of:

• The curve \( y = -x \), a straight line with a negative slope passing through the origin.
• The curve \( y = \sqrt{2-x} \), which is a part of a parabola that opens downwards, bounded by real values of \( y \).
• Two vertical lines: \( x = 0 \) and \( x = 3 \).

Together, these equations define the region's boundaries for integration. Understanding these curves allows the next steps of finding intersections, sketching, and calculating area to become more manageable.

Intersection Points

Intersection points are where the bounding curves meet. To find these, we equate the curves: \( y = -x \) and \( y = \sqrt{2-x} \). Setting \( -x = \sqrt{2-x} \), we square both sides to get \( x^2 = 2 - x \). This rearranges to the quadratic equation \( x^2 + x - 2 = 0 \). Factoring it gives \( (x - 1)(x + 2) = 0 \), resulting in solutions \( x = 1 \) and \( x = -2 \).However, because our region is only between \( x = 0 \) and \( x = 3 \), \( x = -2 \) does not lie within our bounds. Thus, the valid point where these curves intersect is at \( x = 1 \). Plugging this back into either curve gives us \( y = -1 \). The intersection point is therefore \( (1, -1) \). Identifying this helps define the region more accurately for sketching and calculating the area.

Area Calculation

Calculating the area of the region defined by the integral involves evaluating \( \int_{0}^{3} \int_{-x}^{\sqrt{2-x}} dy \, dx \). This integrates over \( y \) first, resulting in \( \int_{-x}^{\sqrt{2-x}} 1 \, dy = \sqrt{2-x} + x \). Afterwards, the calculation continues with respect to \( x \): \( \int_{0}^{3} (\sqrt{2-x} + x) \, dx \).This integral splits into two parts:

• \( \int_{0}^{3} \sqrt{2-x} \, dx \): Substituting \( u = 2-x \), \( du = -dx \), with limits changed from \( u = 2 \) to \( u = -1 \), gives \(-\int_{2}^{0} \sqrt{u} \, du \). The antiderivative \( \frac{2}{3}u^{3/2} \) evaluated gives \( \frac{4\sqrt{2}}{3} \).
• \( \int_{0}^{3} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{3} = \frac{9}{2} \).

Combining these results, the total area is \( \frac{4\sqrt{2}}{3} + \frac{9}{2} \), which completes the calculation for the region's area.

Region Sketching

Sketching the region defined by the double integral translates the mathematical expressions into a visual context. Start by drawing the curves:

• \( y = -x \): This is a straight line through the origin, running from top left to bottom right.
• \( y = \sqrt{2-x} \): Appears as a curve that starts at \( (0, \sqrt{2}) \), travels downward as \( x \) increases, and meets the x-axis as it approaches \( x = 2 \).

The region of interest lies between these two curves and the vertical lines \( x = 0 \) and \( x = 3 \).Finding the intersection point \( (1, -1) \) gives more detail to where these curves cross in the plot, which is where \( y = -x \) negates forcing the parabola shape. This graphical representation allows us to visualize the constraints of the integral, aiding in understanding and confirming the calculated area.