Problem 85 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\iint_{R} x \sec ^{2} y d A ; R=\left\\{(x, y): 0 \leq y \leq x^{2}, 0 \leq x \leq \sqrt{\pi} / 2\right\\}$$

Step-by-Step Solution

Answer

Question: Evaluate the double integral $$\iint_{R} x \sec^{2} y d A$$, where R is the region defined by \((x, y): 0 \leq y \leq x^{2}, 0 \leq x \leq \sqrt{\pi} / 2\). Answer: The double integral is equal to $$\frac{1}{2}\ln(\sqrt{2})$$.

Step 1: Identifying the limits of integration

Given the region R, we can see that the limits of integration for y are from 0 to \(x^{2}\), and for x, the limits are from 0 to \(\sqrt{\pi} / 2\).

Step 2: Evaluate the inner integral

We start by integrating the given function with respect to y: $$\int_0^{x^2} x \sec^2{y} dy$$ Since x is a constant with respect to y, we can take it out of the integral: $$x \int_0^{x^2} \sec^2{y} dy$$ Now, we integrate it with respect to y: $$x\left[ \tan{y} \right]_0^{x^2} = x(\tan(x^2) - \tan(0))$$ The inner integral simplifies to: $$x\tan(x^2)$$

Step 3: Evaluate the outer integral

Now, we have to integrate our result with respect to x: $$\int_0^{\sqrt{\pi}/2} x\tan(x^2) dx$$ To evaluate this integral, we will use substitution. Let \(u = x^2\). Then, \(\frac{du}{dx} = 2x\), or \(dx = \frac{du}{2x}\). The bounds of integration will also change: When \(x = 0\), \(u = 0\). When \(x = \sqrt{\pi} / 2\), \(u = \pi / 4\). Our integral now becomes: $$\int_0^{\pi/4} \tan(u) \frac{du}{2} = \frac{1}{2} \int_0^{\pi/4} \tan(u) du$$

Step 4: Integrating with respect to u

Now we just need to integrate \(\tan(u)\) with respect to u: $$\frac{1}{2} \left[ -\ln(\cos(u)) \right]_0^{\pi/4} = \frac{1}{2} (-\ln(\cos(\pi/4)) + \ln(\cos(0)))$$ We know that \(\cos(\pi/4) = \frac{\sqrt{2}}{2}\) and \(\cos(0) = 1\). So, our result is: $$\frac{1}{2}(-\ln(\frac{\sqrt{2}}{2}) + \ln(1)) = \frac{1}{2}( \ln(2) - \ln(\sqrt{2}))$$

Step 5: Simplifying the result

By using the properties of logarithms, we can simplify the result: $$\frac{1}{2}( \ln(2) - \ln(\sqrt{2})) = \frac{1}{2}(\ln\frac{2}{\sqrt{2}}) = \frac{1}{2}\ln\left(\frac{2\sqrt{2}}{2}\right)$$ $$ = \frac{1}{2}\ln(\sqrt{2})$$ So the integral evaluates to: $$\iint_{R} x \sec^{2} y d A = \frac{1}{2}\ln(\sqrt{2})$$