Problem 52 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

#Short Answer The value of the double integral of x^2y over the region bounded by y=0, y=√x, and y=x-2 is 18.

Step 1: Find the limits of integration

We are given that the region R is bounded by \(y=0, y=\sqrt{x}\) and \(y=x-2\). To find the intersection points of these curves, we will solve the equations pairwise. - Intersection of \(y=0\) and \(y=\sqrt{x}\): \(0 = \sqrt{x} \Rightarrow x=0\) - Intersection of \(y=0\) and \(y=x-2\): \(0 = x-2 \Rightarrow x=2\) - Intersection of \(y=\sqrt{x}\) and \(y=x-2\): \(\sqrt{x}=x-2 \Rightarrow x^2=(x-2)^2 \Rightarrow x=2\) (the result here shows that the two curves are tangent to each other at the intersection point) With these intersection points, we can determine that the region R is bounded by \(x=0, x=2, y=0, y=\sqrt{x}\) and \(y=x-2\).

Step 2: Set up the integral

In order to evaluate the double integral, we need to decide the order of integration. We can choose the order either as \(dA=dxdy\) or \(dA=dydx\). We will choose \(dA=dydx\) since this can be much simpler for this problem. This is because the lower limit of y is always 0 (y-axis) and the upper limit of y is defined by the curve \(y=\sqrt{x}\) or \(y=x-2\). Setting up the integral as follows: $$\iint_R x^2y dA = \int_0^2 \int_0^{\sqrt{x}} x^2y dydx + \int_0^2 \int_0^{x-2} x^2y dydx$$

Step 3: Evaluate the integral

Now we will evaluate the integral step by step. First, we will integrate with respect to y: $$\int x^2y dy = \frac{1}{2}x^2y^2$$ Then we will substitute the limits of integration and sum the two integrals: $$\int_0^2 \left[\frac{1}{2} x^2 (\sqrt{x})^2\right] dx + \int_0^2 \left[\frac{1}{2} x^2 (x-2)^2\right] dx$$ This simplifies to, $$\frac{1}{2} \int_0^2 x^3 dx + \frac{1}{2} \int_0^2 x^4-4x^3+4x^2 dx$$ Next, integrate with respect to x: $$\frac{1}{2} \left[\frac{1}{4}x^4\right]_0^2 + \frac{1}{2} \left[\frac{1}{5}x^5-\frac{1}{x^4}+\frac{4}{3}x^3\right]_0^2$$ Finally, substitute the limits of integration and simplify: $$\frac{1}{2}\left(\frac{1}{4}(2)^4\right)+\frac{1}{2}\left(\frac{1}{5}(2)^5-\frac{1}{(2)^4}+\frac{4}{3}(2)^3\right)-0$$ Simplifying further: $$\frac{1}{2}(4) + \frac{1}{2}\left(\frac{32}{5}-\frac{1}{4}+\frac{32}{3}\right)$$ $$2+\frac{1}{2}\left(\frac{64}{5}+\frac{96}{5}\right)$$ $$2+\frac{160}{10}$$ $$2+16$$ Therefore, the value of the integral is: $$\iint_R x^2y dA = 18$$

Key Concepts

Limits of Integration

Understanding the limits of integration is a vital part of solving double integrals. These limits define the range over which the integration is performed, and they are dependent on the shape of the area we're examining. When we talk about the limits of integration for the double integral \( \iint_{R} x^{2} y d A \), we refer to the precise values of x and y within the region R for which the integral is evaluated.

In this case, the region R is defined by the curves \( y=0 \), \( y=\sqrt{x} \), and \( y=x-2 \). Setting the limits of integration involves finding the intersection points of these curves to create a clearly bounded region. As calculated, the region is defined from \( x=0 \) to \( x=2 \) and the y-values that lie between \( y=0 \) and either \( y=\sqrt{x} \) or \( y=x-2 \) depending on the value of x. This process is crucial because if we set the wrong limits, we can end up evaluating the integral over the wrong area, which would give us an incorrect answer.

Integration Order

The order of integration—whether to integrate with respect to y or x first—is essential for simplifying the process of solving a double integral. This can significantly affect the ease with which the integral can be computed.

In the given problem, choosing \( dA=dydx \) as the order of integration simplifies the calculation. This decision is strategic: because the lower y-bound is always \( y=0 \), and the upper y-bound is either \( y=\sqrt{x} \) or \( y=x-2 \), depending on the x-value. If the order of integration was reversed, setting the limits for x would be more complex, especially since the x-limits are not constants but functions of y in the given region R. Thus, the correct choice of integration order, in this case, \( dydx \) facilitates the problem-solving process.

Area Bounded by Curves

Solving a double integral often involves finding the area bounded by several curves. When visualizing the area bounded by the curves \( y=0 \), \( y=\sqrt{x} \) and \( y=x-2 \), a sketch can be immensely helpful. The process helps in setting up the correct limits for integration.

For the given functions, we can sketch the parabola \( y=\sqrt{x} \) and the line \( y=x-2 \) along with the x-axis (where \( y=0 \) is). The area of interest R is the shape that's formed by these curves. By sketching, we confirm the region we're integrating over, ensuring that we understand the orientation and the points where the curves intersect. In this way, the sketch acts as a visual check which prevents possible errors while setting up our double integral. The integration then sums up all infinitesimal elements (\( dA \)) over this area to give us the total value bounded by these curves.