Question

Sketch the regions of integration and evaluate the following integrals. \(\iint_{R} 3 x y d A ; R\) is bounded by \(y=2-x, y=0,\) and \(x=4-y^{2}\) in the first quadrant.

Short answer

Question: Evaluate the integral \(\iint_{R} 3xy\,dA\), where R is the region enclosed by the line \(y=2-x\), the x-axis, and the parabola \(x=4-y^2\). Answer: The value of the integral is \(\iint_{R} 3xy\,dA = -\frac{23}{2}\).

Step-by-step solution

Sketch the regions

First, let's plot the bounding curves in the first quadrant: - \(y=2-x\): This is a line with a negative slope, which passes through the points \((0,2)\) and \((2,0)\). - \(y=0\): This is the x-axis. - \(x=4-y^2\): This is a parabola opening at the left side, with its vertex at \((4,0)\). Sketch those curves and identify the enclosed region by those three curves, which we will call R.

Determine the order of integration

Since the region R can easily be described vertically, we prefer to use the order of integration as \(dy\,dx\).

Find the bounds of integration

Now let's find the bounds of integration. The x-bound varies from the left parabola to the vertical line given by 2-x, and the y-bound ranges from the x-axis to the intersection of the line \(y=2-x\) and the parabola \(x=4-y^2\). Solve the following system of equations to obtain the intersection point between the two curves: \begin{align*} y &= 2 - x,\\ x &= 4 - y^2. \end{align*} By substitution, we get \(y = 2 - (4-y^2) \Rightarrow y^2+y-2=0\). Solving this quadratic equation for y, we get \(y=1\). Now, substitute this back into the first equation, and we get \(x=1\). Thus, our intersection point is (1,1). Now let's write the bounds of integration: - For x: The x-coordinate varies from the parabola \(x=4-y^2\) to the line \(x=2-y\), or \(4-y^2 \le x \le 2-y\). - For y: The y-coordinate varies from 0 to 1, or \(0 \le y \le 1\).

Evaluate the integral

Now, let's evaluate the integral, given the bounds obtained in step 3: \begin{align*} \iint_{R} 3xy\,dA &= \int_{0}^{1} \int_{4-y^2}^{2-y} 3xy\,dx\,dy \\ &= 3 \int_{0}^{1} \left[\frac{1}{2}x^2y\right]_{4-y^2}^{2-y}\,dy \\ &= 3 \int_{0}^{1} \left[\frac{1}{2}(2-y)^2y - \frac{1}{2}(4-y^2)^2y\right]\,dy \\ &= 3 \int_{0}^{1} \left[\frac{1}{2}(4-4y+y^2)y - \frac{1}{2}(16-8y^2+y^4)y\right]\,dy \\ &= 3 \left[\int_{0}^{1} \frac{1}{2}(4y-4y^2+y^3)\,dy - \int_{0}^{1} \frac{1}{2}(16y-8y^3+y^5)\,dy\right]. \end{align*} Now, we separately integrate with respect to y: \begin{align*} &= 3 \left[\left(\frac{1}{2} \left[\frac{1}{2}y^2-\frac{4}{3}y^3+\frac{1}{4}y^4\right]_{0}^{1} - \left(\frac{1}{2} \left[8y^2-\frac{2}{3}y^4+\frac{1}{6}y^6\right]_{0}^{1}\right)\right] \\ &= 3 \left[\frac{1}{2}\left(\frac{1}{2}-\frac{4}{3}+\frac{1}{4}\right) - \frac{1}{2}\left(8-\frac{2}{3}+\frac{1}{6}\right)\right] \\ &= 3 \left[\frac{1}{2}\left(\frac{9}{12}\right) - \frac{1}{2}\left(\frac{49}{6}\right)\right] \\ &=\frac{3}{4}-\frac{49}{4}=-\frac{46}{4}=-\frac{23}{2}. \end{align*} Therefore, the value of the integral is \(\iint_{R} 3xy\,dA = -\frac{23}{2}\).

Key concepts

Regions of Integration

When working with double integrals, identifying the region of integration is crucial. This involves understanding the area over which the integration occurs. In this exercise, Region R is bounded by three curves: the line \(y = 2 - x\), the x-axis (\(y = 0\)), and the parabola \(x = 4 - y^2\). These lines create a closed shape in the first quadrant of the Cartesian plane.
This region must be sketched to see where the curves intersect and form the boundaries clearly. By visualizing it, we can confirm the enclosed area for integration and ensure our calculations reflect the correct region.

• The line \(y = 2 - x\) is a straight line with a negative slope.
• The parabola \(x = 4 - y^2\) opens to the left.
• The x-axis acts as the lower boundary of the region.

Understanding the shape and boundaries of Region R is the first step towards effectively solving the problem with double integrals.

Bounds of Integration

Setting the bounds of integration correctly is essential for evaluating a double integral over a defined region. The bounds tell us how the variables, in this case, \(x\) and \(y\), change across the region.
In this problem, determining the bounds involves analyzing how each curve acts as a boundary in one direction:

• For the x-coordinate: The integration runs from the parabola \(x = 4 - y^2\) to the line \(x = 2 - y\) due to their positioning.
• For the y-coordinate: The integration begins at \(y = 0\) (the x-axis) and goes up to the line intersection at \(y = 1\).

This creates a clear path for numerical evaluation, ensuring each coordinate follows its limits within the bounded area.

Parabola and Line Intersection

Intersection points of curves define the limits of integration precisely. For double integrals, like in this exercise, identifying these points is essential.
Finding where the parabola \(x = 4 - y^2\) and the line \(y = 2 - x\) intersect involves solving their equations simultaneously. Substituting one equation into the other yields a quadratic equation: \[ y^2 + y - 2 = 0 \]Solving this equation, the intersection turns out to be at \((1, 1)\), confirming one of the integration bounds.
This step ensures the problem's area of integration starts correctly at \((0, 0)\) and adequately captures the overlap of the curves.

Order of Integration

Deciding the order of integration depends on the ease of describing the region and the integration limits. In this exercise, choosing the order \(dy \, dx\) means that we integrate with respect to \(y\) first, then \(x\).
This particular order is advantageous for this region because:

• The bounds for \(y\) are constant, from 0 to 1, making it simple to compute first.
• The bounds for \(x\) rely on \(y\), set between the functions \(4 - y^2\) and \(2 - y\), effectively accommodating the vertical cross-sections of Region R.

By structuring the integration in this direction, we streamline the computation process and effectively handle each area's contribution to the integral. Changing the order would require setting up new bounds, possibly making the problem more complicated.