Question

Evaluate the following integrals. A sketch is helpful. \(\iint_{R} 3 x y d A ; R\) is bounded by \(y=2-x, y=0,\) and \(x=4-y^{2}\) in the first quadrant.

Short answer

Based on the step-by-step solution of evaluating the double integral over the given region R, the value of the integral is -22.

Step-by-step solution

Find the points of intersection

To find the points of intersection, solve the following systems of equations: 1. Intersection of y = 2 - x and y = 0 2. Intersection of y = 2 - x and x = 4 - y^2 3. Intersection of y = 0 and x = 4 - y^2 1. Setting y = 0 in the first equation, x = 2. Therefore, the first point of intersection is (2,0). 2. Setting y = 2 - x in the second equation, we get x = 4 - (2 - x)^2, which simplifies to x^2 - 2x = 0. Solving this quadratic equation, we find that x = 0 or x = 2. Since we are in the first quadrant, x = 0 is not a feasible option, so the second point of intersection is (2, 0). 3. Setting y = 0 in the third equation, we get x = 4, which means the third point of intersection is (4, 0).

Determine the order of integration and limits

The region R is an enclosed area bounded by y = 2 - x (a straight line), y = 0 (the x axis), and x = 4 - y^2 (a parabola). To simplify the integral, we will first integrate with respect to x and then with respect to y. The boundaries for x are determined by the curves y = 2 - x and x = 4 - y^2. The boundaries for y are determined by y = 0 and the parabola x = 4 - y^2. So the integral can be written as follows: $$\int_{0}^{2} \int_{2-y}^{4-y^2} 3xy \,dx\,dy$$

Evaluate the integral with respect to x

Integrate 3xy with respect to x: $$\int_{0}^{2} [ \frac{3}{2}x^{2}y\Big|_{2-y}^{4-y^2} ] \,dy$$ Now substitute the values for the limits of integration: $$\int_{0}^{2} \left[\frac{3}{2}(4-y^2)^{2}y - \frac{3}{2}(2-y)^{2}y\right] \,dy$$

Evaluate the integral with respect to y

Finally, integrate the resulting expression with respect to y: $$\int_{0}^{2} \left[\frac{3}{2}(4-y^2)^{2}y - \frac{3}{2}(2-y)^{2}y\right] \,dy$$ Combining the two terms, we arrive at: $$\frac{3}{2}\int_{0}^{2} \left[(4-y^2)^{2}y - (2-y)^{2}y\right] \,dy$$ Using polynomial expansion and integrating term-by-term, we find: $$\frac{3}{2}\left[\int_{0}^{2} -y^5 + 6y^3 - 15y \, dy \right]$$ $$\frac{3}{2}\left[\left(-\frac{1}{6}y^{6} + \frac{3}{4}y^4 - \frac{15}{2}y^{2}\right) \Big|_0^2\right]$$ $$\frac{3}{2}\left(-\frac{44}{3}\right) = -22$$ The value of the given integral is -22.

Key concepts

Curves of Intersection

In the context of evaluating double integrals, curves of intersection play a critical role in defining the boundaries of the region of integration. In the given exercise, we have three curves:

• The line described by the equation \(y = 2 - x\), which appears as a descending linear path in the first quadrant.
• The horizontal line \(y = 0\), which represents the x-axis.
• The parabola given by \(x = 4 - y^2\), opening towards the left.

Curves of intersection are found where these equations equal each other. For instance, setting \(y = 0\) in the equation \(y = 2 - x\) gives us the point \((2, 0)\). Solving other intersections is essential to find such points, leading us to \((2, 0)\) and \((4, 0)\), all in the first quadrant. These intersections help in visualizing the region \(R\) required for integration.

Integration Limits

Once the region of integration \(R\) is identified through curves of intersection, determining integration limits is the next crucial step in solving a double integral. These limits are vital because they define the area over which the function will be integrated. For the given problem:

• The outer integral over \(y\) runs from \(0\) to \(2\). This range stems from the lowest point on the \(y\)-axis (\(y = 0\)) to where the parabola \(x = 4 - y^2\) intersects the line \(y = 2 - x\).
• The inner integral over \(x\) is bound by the functions \(x = 2 - y\) and \(x = 4 - y^2\). This means \(x\) runs from one boundary to the other depending on \(y\), effectively capturing the entire region \(R\) precisely where it exists within the first quadrant as \(y\) varies.

Assigning these limits correctly is fundamental to covering the desired region accurately. Always ensure every variable in the integral aligns with the physical region described by the curves of intersection.

Order of Integration

Deciding the order of integration involves choosing whether to first integrate with respect to \(x\) or \(y\). This decision impacts the complexity of solving the integral. In the provided example, integrating first with respect to \(x\) is simpler due to the region \(R\) defined by straightforward lines and curves for \(x\) as a function of \(y\).The integration order follows:

• Begin with the inner integral over \(x\), with limits \(2-y\) and \(4-y^2\). Calculate \(\int_{2-y}^{4-y^2} 3xy \,dx\).
• Then, complete the outer integral over \(y\) from \(0\) to \(2\) with the limits determined from the bounds of the region.

Choosing the order of integration can simplify the process significantly, turning potentially complex expressions into more manageable forms. Always aim for clarity and simplicity when deciding the integration order.