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Chapter 16: Problem 14

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

Short Answer

Expert verified
Question: Evaluate the iterated integral $\int_{0}^{\pi / 4} \int_{0}^{3} r \sec \theta d r d \theta.$ Answer: The final value of the iterated integral is $\frac{9}{2} \ln(\sqrt{2} + 1)$.

Step by step solution

01

Identify the iterated integral

The given iterated integral is in the form: $$\int_{0}^{\pi / 4} \int_{0}^{3} r \sec \theta d r d \theta.$$ This represents a double integral where the inner integral, \(I_1\), is being evaluated for \(r\) (going from 0 to 3) and the outer integral, \(I_2\), is being evaluated for \(\theta\) (going from 0 to \(\pi/4\)).
02

Integrate the inner integral

We will first find \(I_1\). Remember that \(\sec\theta\) is a constant with respect to \(r\). Therefore, the inner integral is: $$I_1 = \int_{0}^{3} r \sec \theta d r.$$ Applying power rule: $$I_1 = \left[\frac{r^2}{2} \sec \theta \right]_{0}^{3} = \frac{9}{2} \sec\theta$$
03

Integrate the outer integral

Now we will find \(I_2\), using the result of \(I_1\): $$I_2 = \int_{0}^{\pi / 4} \frac{9}{2} \sec\theta d\theta$$ Integrating \(\sec\theta\) with respect to \(\theta\) gives: $$I_2 = \left[\frac{9}{2} \ln|\sec\theta + \tan\theta|\right]_{0}^{\pi / 4}$$
04

Find the final value

Applying the limits of \(\theta\), we have: $$I_2 = \frac{9}{2} \left[\ln(\sec(\pi / 4) + \tan(\pi / 4)) - \ln(\sec(0) + \tan(0))\right]$$ Remember that \(\sec(\pi / 4) = \sqrt{2}\), \(\tan(\pi / 4) = 1\), \(\sec(0) = 1\), and \(\tan(0) = 0\). Plugging these values in, we get $$I_2 = \frac{9}{2} \left[\ln(\sqrt{2} + 1) - \ln(1) \right]$$ Since \(\ln(1) = 0\), the final value of the iterated integral is $$I_2 = \frac{9}{2} \ln(\sqrt{2} + 1)$$

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Most popular questions from this chapter

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. Let \(R\) be the unit disk centered at \((0,0) .\) Then $$\iint_{R}\left(x^{2}+y^{2}\right) d A=\int_{0}^{2 \pi} \int_{0}^{1} r^{2} d r d \theta$$ b. The average distance between the points of the hemisphere \(z=\sqrt{4-x^{2}-y^{2}}\) and the origin is 2 (calculus not required). c. The integral \(\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} e^{x^{2}+y^{2}} d x d y\) is easier to evaluate in polar coordinates than in Cartesian coordinates.

Find the volume of the solid bounded by the surface \(z=f(x, y)\) and the \(x y\)-plane. (Check your book to see figure) $$f(x, y)=16-4\left(x^{2}+y^{2}\right)$$

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Choose the best coordinate system and find the volume of the following solids. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. The wedge cut from the cardioid cylinder \(r=1+\cos \theta\) by the planes \(z=2-x\) and \(z=x-2\).
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