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: Evaluate the double integral of the function \(y\) over the region R, where R is defined by the inequalities \(0\leq y\leq \sec x\) and \(0\leq x\leq \pi/3\). Answer: The value of the double integral is \(\frac{\pi+1}{6}\).

Step-by-step solution

Set up the integral

First, we set up the integral. We express the given integral as a double integral with the given limits: $$\iint_R y dA = \int_0^{\pi/3}\int_0^{\sec x} y\, dy\, dx$$ This means that we first integrate with respect to \(y\) over the region \([0, \sec x]\), and then integrate with respect to \(x\) over the region \([0, \pi/3]\).

Integrate with respect to y

We will first integrate with respect to \(y\) using the power rule. The power rule states that for a function of the form \(y^n\), the integral is \((y^{n+1})/(n+1)\). \begin{align*} \int_0^{\sec x} y\, dy &= \left[\frac{y^2}{2}\right]_0^{\sec x}\\ &= \frac{(\sec x)^2}{2}-\frac{(0)^2}{2}\\ &= \frac{(\sec x)^2}{2} \end{align*}

Integrate with respect to x

Next, we will integrate the result with respect to \(x\) over the region \([0, \pi/3]\). $$\int_0^{\pi/3} \frac{(\sec x)^2}{2}\, dx$$ To evaluate this integral, we can use a trigonometric identity: \((\sec x)^2 = 1 + (\tan x)^2\). This means that our integral becomes $$\int_0^{\pi/3}\frac{1 + (\tan x)^2}{2}\, dx$$ Now we can split this integral into two parts: \begin{align*} \int_0^{\pi/3}\frac{1 + (\tan x)^2}{2}\, dx &= \frac{1}{2}\int_0^{\pi/3}(1+(\tan x)^2) dx\\ &= \frac{1}{2}\int_0^{\pi/3} 1\, dx + \frac{1}{2}\int_0^{\pi/3} (\tan x)^2\, dx \end{align*}

Evaluate the integrals

Now we can evaluate the remaining integrals: \begin{align*} \frac{1}{2}\int_0^{\pi/3} 1\, dx &= \frac{1}{2} \left[x\right]_0^{\pi/3}\\ &= \frac{1}{2} \pi/3 =\frac{\pi}{6} \end{align*} For the second part, we can use substitution. Let \(u = \tan x\), which means \(du = (\sec x)^2 dx\). Our integral becomes: $$\frac{1}{2}\int_0^{\pi/3} (\tan x)^2\, dx = \frac{1}{2}\int_0^1 u^2\, du$$ Now, we can use the power rule again to evaluate this integral: \begin{align*} \frac{1}{2}\int_0^1 u^2\, du &= \frac{1}{2}\left[\frac{u^3}{3}\right]_0^1\\ &= \frac{1}{6} \end{align*}

Combine the results

Now we can combine the results: \begin{align*} \iint_R y dA &= \frac{\pi}{6} + \frac{1}{6}\\ &= \frac{\pi+1}{6} \end{align*} The value of the double integral is \(\frac{\pi+1}{6}\).