Question

Sketch the regions of integration and evaluate the following integrals. \(\iint_{R} 12 y d A ; R\) is bounded by \(y=2-x, y=\sqrt{x},\) and \(y=0\)

Short answer

In conclusion, the double integral of the given function over the region R is equal to 16. We first sketched the functions and found their points of intersection to determine the limits of integration, then we set up and evaluated the double integral to find the final result.

Step-by-step solution

Sketch the functions and find the points of intersection

To begin, we need to sketch the functions \(y = 2 - x, y = \sqrt{x},\) and \(y = 0\) on the \(xy\)-plane. Then, find where these functions intersect each other to get the boundaries of the region R.

Set up the integral

Now that we have the region R, we can set up the double integral. We will integrate first with respect to \(dx\) and then with respect to \(dy\). To do this, we need to express the \(x\) limits in terms of \(y\). We have the following equations: \(y = 2 - x\) can be re-arranged as \(x = 2 - y\) \(y = \sqrt{x}\) can be re-arranged as \(x = (y^2)\) The limits of integration for \(y\) are \(0\) and \(2\), since these are the minimum and maximum \(y\) values in the region R. As we found in Step 1, the limits of integration for \(x\) go from \((y^2)\) to \((2 - y)\). Therefore, we set up the integral as follows: $$\iint_{R} 12y\,dA = \int_{0}^{2} \int_{y^2}^{2 - y} 12y\,dxdy$$

Evaluate the integral

Now that we have our integral set up, we can proceed with evaluating it. First, we will integrate the inner integral with respect to \(x\): $$\int_{y^2}^{2 - y} 12y\,dx = 12y \left[\int_{y^2}^{2 - y} dx\right] = 12y[x]_{y^2}^{2 - y} = 12y[(2 - y) - (y^2)] = 12y(2 - y) - 12y^3$$ Now, integrate the outer integral with respect to \(y\): $$\int_{0}^{2} (12y(2 - y) - 12y^3)dy = \int_{0}^{2} (24y - 12y^2 - 12y^3)dy = [12y^2 - 4y^3 - 3y^4]_{0}^{2}$$ Plugging in the values of \(y = 2\) and \(y = 0\), we get: $$[12(2)^2 - 4(2)^3 - 3(2)^4] - [0] = [48 - 32 - 48] = 16$$ The value of the integral is \(16\).

Key concepts

Regions of Integration

When dealing with double integrals, one crucial step is identifying the region over which you are integrating. This region, known as the region of integration, is a specific area on the coordinate plane that is bounded by the graphs of the given equations. In our example, the functions that form this region are:

• The line: \(y = 2 - x\), a straight line sloping downward where y decreases by 1 as x increases by 1.
• The curve: \(y = \sqrt{x}\), a simple parabolic curve that resembles the right half of a much flattened U-shape.
• The line: \(y = 0\), which is the horizontal x-axis.

To find the region of integration, sketch these functions on the \(xy\)-plane. The intersections of these graphs delineate the boundaries of the region R. For this problem, solving these intersections gives us the critical points that will define our integration limits. In practice, this means finding where the curves intersect each other: - The intersection of \(y=2-x\) and \(y=\sqrt{x}\) gives the point (1,1). - The intersection of \(y=\sqrt{x}\) and the x-axis \(y=0\) gives the point (0,0). - The intersection for the line \(y=2-x\) and the x-axis \(y=0\) gives the point (2,0). These points create a closed boundary creating a region in the shape of a triangle and a curved boundary situated in the first quadrant. Understanding the layout and border of this region allows us to accurately set up the limits for our integrals.

Evaluating Integrals

Once we have sketched the region of integration, the next step is to evaluate the double integral over this region. This involves setting up and calculating an integral expression with limits defined by the region's boundaries. For our example, the double integral \[ \iint_{R} 12y \, dA = \int_{0}^{2} \int_{y^2}^{2 - y} 12y \, dx \, dy \] represents this process. Here's a breakdown of how to evaluate it:**Step 1: The Inner Integral** First, integrate with respect to \(x\). For this, our bounds are from \(x = y^2\) to \(x = 2-y\). The expression arises from rewriting the original equations to solve for \(x\) in terms of \(y\). The integration yields:\[\int_{y^2}^{2 - y} 12y \, dx = 12y[(2-y) - (y^2)] = 12y(2-y-y^2) \]**Step 2: The Outer Integral** Next, integrate the result from Step 1 with respect to \(y\) from 0 to 2:\[\int_{0}^{2} (24y - 12y^2 - 12y^3) \, dy \]This evaluates to:\[[12y^2 - 4y^3 - 3y^4]_{0}^{2} \]By plugging in the limits, the final result is calculated. This process of integrating step by step, first with respect to one variable and then with respect to the next, simplifies our task of calculating complex multi-variable integrals.

Boundaries of Regions

Establishing the boundaries of the region is essential for accurately setting up double integrals. In our example, these boundaries corresponded to the lines and curves: \(y = 2-x\), \(y = \sqrt{x}\), and \(y = 0\). These equations intersect each other at specific points, which serve as limits for integration:

• From \(y = 2-x\) and \(y = \sqrt{x}\), we found the intersection point (1,1).
• From \(y = \sqrt{x}\) and \(y = 0\), the intersection is (0,0), where the curve meets the x-axis.
• From \(y = 2-x\) and the x-axis \(y = 0\), the intersection is (2,0).

Within these boundaries, the area R resembles a characteristic triangular region with a curvilinear side. The conventions for the region of integration involve these points and dictate both the internal and external limits for the integral. To set these limits, imagine the region projected onto each axis. When integrating, consider the entirety of the region. Here, \(y\) values range from 0 to 2, while specific \(x\) values range from \(y^2\) to \(2-y\). This approach in determining the bounds ensures that your calculations fully encompass the area described by the given equations.