Problem 16 - Calculus Early Transcendentals

Question

Evaluate the following iterated integrals. $$\int_{0}^{\pi / 4} \int_{0}^{3} r \sec \theta d r d \theta$$

Step-by-Step Solution

Answer

Based on the step by step solution above, evaluate the iterated integral $$\int_{0}^{\frac{\pi}{4}} \int_{0}^{3} r \sec \theta \,dr\,d\theta$$ Solution: The numerical value of the iterated integral is $$\frac{9}{2} \ln(\sqrt{2}+1)$$

Step 1: Identify the order of integration

The given iterated integral is in the order of integration \(\iint d rd\theta\). The first integral will be evaluated with respect to \(r\), and the second integral will be evaluated with respect to \(\theta\).

Step 2: Evaluate the integral with respect to \(r\)

Let's evaluate the inner integral: $$\int_{0}^{3} r \sec \theta dr$$ To evaluate this integral, we can rewrite it as: $$\sec \theta \int_{0}^{3} r dr$$ Since \(\sec \theta\) is constant with respect to \(r\), we can take it outside the integral. Now, integrate \(r\) with respect to \(r\): $$\sec \theta \left[ \frac{1}{2} r^2 \right]_0^3$$ Now, substitute the limits to get: $$\sec \theta \left[ \frac{1}{2} (3^2 - 0^2) \right] = \frac{9}{2} \sec \theta$$

Step 3: Substitute the result of step 2 into the original integral

Now that we have the result from step 2, which is \(\frac{9}{2} \sec \theta\), we substitute this into the outer integral: $$\int_{0}^{\frac{\pi}{4}} \frac{9}{2} \sec \theta d\theta$$

Step 4: Evaluate the integral with respect to \(\theta\)

Now, we need to evaluate the integral with respect to \(\theta\): $$\int_{0}^{\frac{\pi}{4}} \frac{9}{2} \sec \theta d\theta$$ We can take out the constant term \(\frac{9}{2}\) and rewrite the integral as: $$\frac{9}{2} \int_{0}^{\frac{\pi}{4}} \sec \theta d\theta$$ Now, integrate \(\sec \theta\) with respect to \(\theta\): $$\frac{9}{2} \left[ \ln | \sec \theta + \tan \theta | \right]_0^{\frac{\pi}{4}}$$

Step 5: Substitute the limits into the integral

Now, substitute the limits of integration to find the final answer: $$\frac{9}{2} \left[ \ln | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} | - \ln | \sec 0 + \tan 0 | \right]$$ Now, simplify to get: $$\frac{9}{2} \left[ \ln | \sqrt{2} + 1 | - \ln | 1 | \right]$$ Finally, simplify the expression to get the final answer: $$\boxed{\frac{9}{2} \ln(\sqrt{2}+1)}$$