Problem 42 - Calculus Early Transcendentals

Question

Evaluate the following integrals in spherical coordinates. $$\int_{0}^{2 \pi} \int_{0}^{\pi / 3} \int_{0}^{4 \sec \varphi} \rho^{2} \sin \varphi d \rho d \varphi d \theta$$

Step-by-Step Solution

Answer

Question: Evaluate the triple integral in spherical coordinates: $$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{0}^{4\sec\varphi} \rho^{2}\sin\varphi\ d\rho\ d\varphi\ d\theta$$ Answer: The triple integral evaluates to $$\frac{128\pi}{9}.$$

Step 1: Understanding the limits of integration

The limits of integration are given in spherical coordinates. Keep in mind the order of integration that has been specified: $$d\rho\ d\varphi\ d\theta.$$ The limits for each variable are: - \(\theta\) ranges from \(0\) to \(2\pi\). - \(\varphi\) ranges from \(0\) to \(\frac{\pi}{3}\). - \(\rho\) ranges from \(0\) to \(4\sec\varphi\).

Step 2: Integrate with respect to \(\rho\)

Now we will integrate the integrand with respect to \(\rho\). $$\int_{0}^{4\sec\varphi} \rho^{2} \sin\varphi d\rho = \frac{1}{3}\rho^3\sin\varphi \Big|_{0}^{4\sec\varphi} = \frac{1}{3} (4\sec\varphi)^3\sin\varphi$$ This simplifies to: $$\frac{64}{3}\sec^3\varphi\sin\varphi$$ Now our integral becomes: $$\int_{0}^{2\pi} \int_{0}^{\pi/3} \frac{64}{3}\sec^3\varphi\sin\varphi d\varphi d\theta$$

Step 3: Integrate with respect to \(\varphi\)

Next, we will integrate the integrand with respect to \(\varphi\): $$\int_{0}^{\pi/3} \frac{64}{3} \sec^3\varphi \sin\varphi d\varphi$$ We will use the substitution method, with \(u=\cos\varphi,\) so that \(du=-\sin\varphi d\varphi.\) Then the integral becomes: $$-\frac{64}{3}\int_{1}^{\frac{1}{2}} u^{-3} du = \frac{64}{3}\int_{1}^{\frac{1}{2}}u^{-3}du$$ Now we can integrate with respect to \(u\): $$\frac{64}{3}\int_{1}^{\frac{1}{2}} u^{-3} du = \frac{64}{3}\left[\frac{1}{2}u^{-2}\right]_1^{\frac{1}{2}} = \frac{32}{3}\left(\left(\frac{1}{2}\right)^{-2} - 1\right)$$ After simplification, this evaluates to: $$\frac{64}{9}$$ Now our integral becomes: $$\int_{0}^{2\pi} \frac{64}{9} d\theta$$

Step 4: Integrate with respect to \(\theta\)

Finally, we will integrate the integrand with respect to \(\theta\): $$\int_{0}^{2\pi} \frac{64}{9} d\theta = \frac{64}{9}\theta \Big|_{0}^{2\pi} = \frac{64}{9}(2\pi)$$

Step 5: Final result

After integrating with respect to all three variables, our final result is: $$\frac{64}{9}(2\pi) = \boxed{\frac{128\pi}{9}}$$

Key Concepts

Triple Integral

The concept of a triple integral extends the idea of a double integral to functions of three variables. In a triple integral, we're finding the volume under a surface within a three-dimensional region. To perform a triple integral in spherical coordinates, we express the volume element as \(dV = \rho^2 \sin\varphi\, d\rho\, d\varphi\, d\theta\). The order of the integration—\(d\rho\, d\varphi\, d\theta\)—is important as it designates the sequence in which integrations with respect to the variables will occur. The limits of integration often describe a spherical region or part of it.

Triple integrals are used in various fields such as physics, engineering, and probability to calculate mass, density, and other spatial properties of three-dimensional objects.

Spherical Coordinates

Spherical coordinates \( (\rho, \varphi, \theta) \) are a system for representing points in three-dimensional space, where \(\rho\) is the radius or the distance from the origin, \(\varphi\) is the polar angle measured from the positive z-axis, and \(\theta\) is the azimuthal angle in the xy-plane from the positive x-axis.

Using spherical coordinates can simplify integrals that have spherical symmetry. In our exercise, the spherical symmetry of the integrand and the limits of integration allows the use of spherical coordinates, which are particularly advantageous for this problem since the limits are simple multiples of π and the integrand evenly distributes over the spherical surface.

Integration Techniques

Successfully evaluating complex integrals often requires the use of advanced integration techniques. Such techniques may include substitution, integration by parts, trigonometric integrals, partial fraction decomposition, and others.

In our problem, the initial integration is straightforward, but as we progress, we encounter a \( \sec^3\varphi \) term. This is where integration techniques such as substitution prove vital. We use the substitution method to solve for \( \varphi \) by relating it to \(u = \cos\varphi\), which simplifies the integral into a more manageable form. Knowing multiple integration techniques and recognizing when to apply them is essential for solving complex integrals efficiently.

Substitution Method

The substitution method, also known as u-substitution, is a tool for integrating functions that involve composite functions, which can be unwieldy to integrate in their original form. By choosing \( u \) as a function of \( x \) (in our case \( u = \cos(\varphi) \) and thus \(du = -\sin(\varphi) d\varphi\)), we transform the integral into one in terms of \( u \) with hopefully simpler limits and integrand.

In the example problem, after performing the substitution, we encounter \( u^{-3} \) rather than \( \sec^3\varphi \)—a much simpler expression to integrate. Correctly applying the substitution method is key to finding solutions for certain kinds of integrals, as it often turns an intractable problem into a textbook example of basic integration.