Question

Evaluate the following integrals. $$\iint_{R} y d A ; R=\\{(x, y): 0 \leq y \leq \sec x, 0 \leq x \leq \pi / 3\\}$$

Short answer

Question: Calculate the double integral of the function y over the region given by the conditions \(0 \leq y \leq \sec x\) and \(0 \leq x \leq \frac{\pi}{3}\). Answer: The double integral of the function y over the given region is \(\frac{16}{27}\).

Step-by-step solution

Identify the limits of integration

The given region R is a sector with the conditions \(0 \leq y \leq \sec x\) and \(0 \leq x \leq \frac{\pi}{3}\). We will integrate over this region.

Change the integral into polar coordinates

We first rewrite the given function and limits using polar coordinates. For any given point \((x,y)\) we have, \(x = r\cos\theta\), \(y = r\sin\theta\) and \(dA = r\, dr\, d\theta\). So, our integral becomes: $$\iint_{R} y d A = \iint_{R} (r\sin\theta) r\, dr\, d\theta$$Now, we also need to find the limits for r and θ.

Find the limits for r and θ

By the conditions \(0 \leq y \leq \sec x\) and \(0 \leq x \leq \frac{\pi}{3}\), it's easy to see that, \(\theta \in [0, \frac{\pi}{3}]\). Now for the condition \(y \leq \sec x\), we write it in terms of polar coordinates as, \(r\sin\theta \leq \frac{1}{\cos\theta}\), which implies, \(r\leq \frac{\sin\theta}{\cos^2\theta}\). So, the limits are \(0 \leq r \leq \frac{\sin\theta}{\cos^2\theta}\) and \( 0 \leq \theta \leq \frac{\pi}{3}\).

Compute the integral

Now compute the integral: \begin{align*} \iint_{R} y d A &= \int_{\theta=0}^{\frac{\pi}{3}}\int_{r=0}^{\frac{\sin\theta}{\cos^2\theta}} (r\sin\theta)r \, dr \, d\theta \\ &= \int_{\theta=0}^{\frac{\pi}{3}} (\sin\theta) \int_{r=0}^{\frac{\sin\theta}{\cos^2\theta}} r^2 \, dr \, d\theta \\ \end{align*} We first integrate with respect to r: $$\int_{r=0}^{\frac{\sin\theta}{\cos^2\theta}}r^2\, dr= \left[\frac{r^3}{3}\right]_{0}^{\frac{\sin\theta}{\cos^2\theta}} = \frac{\sin^3\theta}{3\cos^6\theta}$$

Integrate with respect to θ

Now, we integrate the remaining integral with respect to θ: \begin{align*} \int_{\theta=0}^{\frac{\pi}{3}} \frac{\sin^3\theta}{3\cos^6\theta}d\theta&=\frac{1}{3}\int_{\theta=0}^{\frac{\pi}{3}}\frac{\sin^3\theta}{\cos^6\theta}\, d\theta \end{align*} This integral requires a different approach to evaluate. We will apply the strategy, \(\sin^3\theta = (\sin^2\theta)(\sin\theta) = (1-\cos^2\theta)(\sin\theta)\), and then change the variable to \(u = \cos\theta\) so we have \(du = -\sin\theta d\theta\). \begin{align*} \frac{1}{3}\int_{\theta=0}^{\frac{\pi}{3}}\frac{\sin^3\theta}{\cos^6\theta}\, d\theta &= \frac{1}{3}\int_{\theta=0}^{\frac{\pi}{3}}\frac{(1-\cos^2\theta)(\sin\theta)}{\cos^6\theta} d\theta \\ &= \frac{-1}{3}\int_{\theta=0}^{\frac{\pi}{3}}\frac{(1-\cos^2\theta)}{\cos^5\theta}(-\sin\theta) d\theta \\ &= \frac{-1}{3}\int_{u=1}^{\frac{1}{2}}\frac{(1-u^2)}{u^5} du \\ \end{align*} Evaluate this definite integral. $$\frac{-1}{3}\int_{u=1}^{\frac{1}{2}}\frac{(1-u^2)}{u^5} du = \frac{-1}{3}\left[6u + \frac{4}{3}u^{-3} - u^{-5}\right]_{\frac{1}{2}}^{1}$$

Final result

Now, we can plug in the limits and find the result: $$\frac{-1}{3}\left[6u + \frac{4}{3}u^{-3} - u^{-5}\right]_{\frac{1}{2}}^{1} = \frac{-1}{3}\left(\left[6\cdot\frac{1}{2} + \frac{4}{3}\left(\frac{1}{2}\right)^{-3} - \left(\frac{1}{2}\right)^{-5}\right] - \left[6 + \frac{4}{3} - 1\right]\right)=\frac{16}{27}$$ Thus, the final result of the double integral is: $$\iint_{R} y d A=\frac{16}{27}.$$

Key concepts

Polar Coordinates

Polar coordinates provide a way to express points in the plane using a distance and angle instead of the usual x and y coordinates. By using

• \( x = r\cos\theta \)
• \( y = r\sin\theta \)

we convert the integral into a format that often simplifies problems with circular symmetry.

Here, \( r \) is the distance from the origin, and \( \theta \) is the angle from the positive x-axis. This conversion is especially helpful when dealing with circular regions or angles, where Cartesian coordinates can be cumbersome.

For integration, the area element \( dA \) becomes \( r \, dr \, d\theta \), accounting for the way areas are measured in polar coordinates. This also helps in setting up and evaluating integrals more intuitively in circular or radial problems.

Limits of Integration

Setting correct limits of integration is crucial for evaluating multiple integrals. These limits define the region of integration in the plane.

In our example:

• \( 0 \leq \theta \leq \frac{\pi}{3} \) defines the angular extent.
• \( 0 \leq r \leq \frac{\sin\theta}{\cos^2\theta} \) defines the radial extent.

Transferring these limits from Cartesian to polar coordinates helps to specify regions that have circular boundaries or are defined by angles. Understanding the graphical representation of these limits is as critical as the algebraic method to grasp the extent of integration.

Integration by Substitution

Integration by substitution is a powerful technique to simplify integrals. By changing variables, we can turn a complex integral into a more manageable one.

In this problem, substituting \( \sin\theta = (1 - \cos^2\theta) \sin\theta \) was key, leading to the variable change \( u = \cos\theta \), thereby simplifying the integral drastically.

This method often employs derivatives like \( du = -\sin\theta d\theta \), which help in transforming the integral's limits and its differential part. The process underscores the importance of finding the right substitution to exploit symmetry or simplicity in the original problem.

Volume Calculation

Double integrals can be used to calculate volumes under surfaces. The integral we evaluated represents the volume under the surface \( y = \sec x \) over the region \( R \).

This application requires understanding:

• How the surface in question relates to the function being integrated.
• The role of limits in defining the precise region beneath the surface.

By translating the area integral into polar coordinates and integrating correctly, we obtain the volume directly. This illustrates how multidimensional integrals extend the concept of area to higher dimensions, serving both geometric and physical applications in calculating volumes.