Problem 26 - Thomas Calculus

Question

Evaluate the spherical coordinate integrals. \(\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{\sec \phi}(\rho \cos \phi) \rho^{2} \sin \phi d \rho d \phi d \theta\)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{\pi}{4}\).

Step 1: Understand the Integral

The given integral is in spherical coordinates. We need to evaluate it over the region defined by the limits: - \(\theta\) from \(0\) to \(2\pi\),- \(\phi\) from \(0\) to \(\pi/4\),- \(\rho\) from \(0\) to \(\sec\phi\).The integrand is \(\rho \cos\phi \rho^2 \sin\phi=\rho^3 \cos\phi \sin\phi\). This product must be evaluated over these boundaries.

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

First, integrate \(\rho^3 \cos\phi \sin\phi\) with respect to \(\rho\) from \(0\) to \(\sec\phi\) while treating \(\phi\) and \(\theta\) as constants:\[\int_{0}^{\sec \phi} \rho^3 \cos \phi \sin \phi \, d \rho = \cos \phi \sin \phi \left[ \frac{\rho^4}{4} \right]_{0}^{\sec \phi} = \cos \phi \sin \phi \cdot \frac{(\sec \phi)^4}{4}.\]Note that \((\sec\phi)^4 = \frac{1}{\cos^4\phi}\).

Step 3: Simplify the Expression

Simplify the expression from the previous step:\[\frac{1}{4} \cos \phi \sin \phi \cdot \frac{1}{\cos^4 \phi} = \frac{1}{4} \sin \phi \cdot \frac{1}{\cos^3 \phi} = \frac{1}{4} \tan \phi \sec^3 \phi.\]

Step 4: Integrate with Respect to \(\phi\)

Now, integrate the result \(\frac{1}{4} \tan\phi \sec^3\phi\) with respect to \(\phi\) from \(0\) to \(\pi/4\):\[\int_{0}^{\pi/4} \frac{1}{4} \tan \phi \sec^3 \phi \, d \phi.\]Use the identity \(\frac{d}{d\phi}(\sec\phi)=\sec\phi\tan\phi\) and recognize an integral form:\[\frac{1}{4} \int \sec \phi \tan \phi \cdot \sec^2 \phi \; d \phi = \frac{1}{4} \, \left[ \frac{1}{2} \sec^2 \phi \right]_{0}^{\pi/4}.\]

Step 5: Evaluate the Antiderivative

Evaluate \(\frac{1}{4} \cdot \frac{1}{2} [\sec^2 \phi]_{0}^{\pi/4}\):\[\frac{1}{8} \left( [\sec^2 (\pi/4)] - [\sec^2 (0)] \right) = \frac{1}{8} (2 - 1) = \frac{1}{8}.\]

Step 6: Integrate with Respect to \(\theta\)

Finally, integrate the result \(\frac{1}{8}\) with respect to \(\theta\) over \([0, 2\pi]\):\[\int_{0}^{2\pi} \frac{1}{8} \, d \theta = \frac{1}{8} \cdot \theta \bigg|_0^{2\pi} = \frac{1}{8} \cdot (2\pi) = \frac{\pi}{4}.\]

Step 7: Conclusion

Combine all the results to conclude that the value of the integral is \(\frac{\pi}{4}\).

Key Concepts

Triple Integral

A triple integral is a powerful tool in calculus used to calculate the volume under a three-dimensional region. This involves integrating a function over three different variables. In this context, the variables are \( \rho \), \( \phi \), and \( \theta \) in spherical coordinates.

• The idea of a triple integral is similar to that of single and double integrals, but it extends to three dimensions.
• It allows us to accumulate quantities like mass or charge density over a volume.
• We perform the integration iteratively with respect to each variable in sequence.

In the provided problem, we start by fixing two variables and integrating with respect to the third one. Each integral adopts its limit from the geometric properties of the sphere.
Understanding and setting these limits is critical, as it defines the region of space being considered. Through this approach, the triple integral transforms complex dimensional data into a comprehensible result.

Spherical Coordinates

Spherical coordinates are another way to represent points in three-dimensional space, particularly useful for dealing with spheres or spherical objects.

• These coordinates are represented by \( \rho \), \( \theta \), and \( \phi \).
• Here, \( \rho \) is the radial distance from the origin (position in space), \( \theta \) is the azimuthal angle in the xy-plane from the x-axis, and \( \phi \) is the polar angle.
• Spherical coordinates simplify the handling of symmetries around a sphere, making them perfect for volume integrals involving spherical regions.

The exercise provided uses these coordinates to ease the integration over a specific region constrained by the angles \( \phi \) and \( \theta \), and by the radial distance \( \rho \). The transformation from Cartesian to spherical coordinates requires adjustments to the integrand, such as multiplying by \( \rho^2 \sin \phi \), which accounts for the volume element in spherical coordinates. This transformation is key to setting up the integral correctly.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of more than one variable. It helps analyze complex systems where change occurs over different dimensions, which is essential for understanding real-world phenomena.

• Key concepts include partial derivatives, gradient, divergence, and integrals across multiple dimensions.
• Triple integrals, like in this problem, are a direct application of multivariable calculus.
• They require an understanding of how to integrate functions over domains defined by other variables.

In this problem, multivariable calculus is at the core as it involves integrating over a spherical region. You have to be adept in handling the changes between different coordinate systems, especially when moving from Cartesian to spherical coordinates.
This broadens your ability to tackle diverse problems, from physics to engineering, where such integrals are common.