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Chapter 15: Problem 5

Evaluate the iterated integral. $$\int_{0}^{\pi / 2} \int_{0}^{\cos \theta} e^{\sin \theta} d r d \theta$$

Short Answer

Expert verified
The value of the integral is \(e - 1\).

Step by step solution

01

Understanding the Iterated Integral

The iterated integral is given as: \(\int_{0}^{\pi / 2} \int_{0}^{\cos \theta} e^{\sin \theta} \ dr \ d\theta\). We need to integrate with respect to \(r\) first (inner integral) and then with respect to \(\theta\) (outer integral). Notice that \(e^{\sin \theta}\) is a constant with respect to \(r\).
02

Evaluating the Inner Integral

Integrate \(e^{\sin \theta}\) with respect to \(r\) over \([0, \cos \theta]\):\[\int_{0}^{\cos \theta} e^{\sin \theta} \ dr = e^{\sin \theta} \times r \bigg|_{0}^{\cos \theta} = e^{\sin \theta} \times (\cos \theta - 0) = e^{\sin \theta} \cos \theta.\]The result of the inner integral is \(e^{\sin \theta} \cos \theta\).
03

Evaluating the Outer Integral

Next, integrate the result with respect to \(\theta\) from \(0\) to \(\frac{\pi}{2}\):\[\int_{0}^{\pi / 2} e^{\sin \theta} \cos \theta \ d\theta.\]Notice that the derivative of \(\sin \theta\) is \(\cos \theta\), so we can use substitution: Let \( u = \sin \theta \), then \( du = \cos \theta \ d\theta \). When \(\theta = 0\), \(u = 0\), and when \(\theta = \frac{\pi}{2}\), \(u = 1\).
04

Solving the Integral Using Substitution

With substitution, the integral becomes:\[\int_{0}^{1} e^u \ du.\]The integral of \(e^u\) is simply \(e^u\):\[e^u \bigg|_{0}^{1} = e^1 - e^0 = e - 1.\]
05

Conclusion

Thus, the value of the original iterated integral is \(e - 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multiple Integrals
When we talk about multiple integrals, we're extending the idea of a single integral to more than one dimension. Essentially, they help us integrate functions over a region in two or more steps. In this context, we're dealing with a double integral. It involves integrating with respect to one variable first, then another.

Here, our iterated integral \[\int_{0}^{\pi / 2} \int_{0}^{\cos \theta} e^{\sin \theta} \, dr \, d\theta\]means:
  • We'll integrate over the variable \(r\) first, using the inner integral \(\int_{0}^{\cos \theta} e^{\sin \theta} \, dr \).
  • Then, we'll perform an integration over \(\theta\) with the outer integral \(\int_{0}^{\pi / 2} \, d\theta\).
This sequence of operations allows us to sum small elements across a region described by the limits. The outcome gives a cumulative sum that represents the area or volume under the surface defined by the function.
Integration Techniques
There are a lot of integration techniques available, each suitable for different scenarios. Here we've used a straightforward technique for the inner integral because its integrand, \(e^{\sin \theta}\), is constant with respect to \(r\). This means the integration behaves like multiplying the constant with the range of \(r\).

To solve:
  • We first take \(e^{\sin \theta}\) outside the integral since it's not dependent on \(r\).
  • Integrate \(r\) over the interval \([0, \cos \theta]\).
You'll end up with \(e^{\sin \theta} \times (\cos \theta - 0)\). This is a handy method when part of the function is separable. The simplicity of the inner integral nicely utilizes this basic multiplication technique.
Substitution Method
The substitution method is a powerful tool in calculus. It's like changing gears to simplify integration when facing a challenging function. For the outer integral in our problem \[\int_{0}^{\pi / 2} e^{\sin \theta} \cos \theta \, d\theta,\]substitution shines because of the relationship between sine and cosine.

Here's how it works:
  • We substitute \(u = \sin \theta\), which makes \(du = \cos \theta \, d\theta\).
  • This transforms the integral into \(\int_{0}^{1} e^{u} \, du\).
  • With simpler bounds \(u: 0 \rightarrow 1\), integration becomes next to effortless, yielding \(e^{u} \bigg|_{0}^{1}\).
The result is easy to compute: \(e - 1\). Using substitution cuts down complexity, turning trigonometrical expressions into straightforward exponential ones.

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Most popular questions from this chapter

Express the integral \(\iint_{E} f(x, y, z) d V\) as an iterated integral in six different ways, where \(E\) is the solid bounded by the given surfaces. $$ y^{2}+z^{2}=9, \quad x=-2, \quad x=2 $$

Assume that the solid has constant density \(k\) Find the moment of inertia about the \(z\) -axis of the solid cylinder \(x^{2}+y^{2} \leqslant a^{2}, 0 \leqslant z \leqslant h .\)

Use cylindrical coordinates. Find the volume of the solid that lies within both the cylinder \(x^{2}+y^{2}=1\) and the sphere \(x^{2}+y^{2}+z^{2}=4\)

Sketch the region of integration and change the order of integration. $$\int_{0}^{1} \int_{4 x}^{4} f(x, y) d y d x$$

\(21-34\) Use spherical coordinates. Evaluate \(\iiint_{E} e^{\sqrt{x^{2}+y^{2}+z^{2}}} d V,\) where \(E\) is enclosed by the sphere \(x^{2}+y^{2}+z^{2}=9\) in the first octant.
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