Question

Evaluate the following integrals in spherical coordinates. \(\int_{0}^{\pi} \int_{0}^{\pi / 6} \int_{2 \sec \varphi}^{4} \rho^{2} \sin \varphi d \rho d \varphi d \theta\)

Short answer

Question: Evaluate the triple integral of the function \(\rho^2 \sin \varphi\) over the region bounded by \(2 \sec \varphi \le \rho \le 4, 0 \le \varphi \le \pi/6, \) and \(0 \le \theta \le \pi\) in spherical coordinates. Answer: \(\pi\left(64(1-\cos(\pi/6)) + \frac{14}{3}\right)\)

Step-by-step solution

Integration with respect to \(\rho\)

The first step is to integrate the given function with respect to \(\rho\). The integral with respect to \(\rho\) is : \(\int_{2 \sec \varphi}^{4} \rho^2 d\rho\) This is a simple power rule integration problem, so the result is: \(\frac{1}{3} (\rho^3 \Big|_{2 \sec \varphi}^{4}) = \frac{1}{3} (64- 8 \sec^3 \varphi)\) So, our current integral becomes: \(\int_{0}^{\pi} \int_{0}^{\pi / 6} \frac{1}{3} (64- 8 \sec^3 \varphi) \sin \varphi d \varphi d \theta\)

Integration with respect to \(\varphi\)

Now, we need to integrate with respect to \(\varphi\): \(\int_{0}^{\pi / 6} \frac{1}{3} (64- 8 \sec^3 \varphi) \sin \varphi d \varphi\) This integral might be a bit more difficult, so let's break it into two parts. \(\frac{1}{3} \int_{0}^{\pi / 6} 64 \sin \varphi d \varphi - \frac{8}{3} \int_{0}^{\pi / 6} \sec^3 \varphi \sin \varphi d \varphi\) The first integral is simple to evaluate: \( \frac{1}{3} \int_{0}^{\pi / 6} 64 \sin \varphi d \varphi = 64 (-\cos \varphi \Big|_{0}^{\pi / 6})\) \(=64(1-\cos(\pi/6))\). For the second integral, notice that \(\sin \varphi = \frac{\sec \varphi}{\sec \varphi}\), so we can rewrite the integral as follows: \(-\frac{8}{3} \int_{0}^{\pi / 6} \sec^4 \varphi d(\cos \varphi)= -\frac{8}{3}\left(-\frac{\sec^4 \varphi\cos \varphi}{4}\Big|_{0}^{\pi/6}\right) = \frac{2}{3}(\sec^4 (\pi / 6) - \sec^4 (0)) = \frac{2}{3}((2^4) - (1^4)) = \frac{14}{3}\) Now, our current integral becomes: \(\int_{0}^{\pi} 64(1-\cos(\pi/6)) + \frac{14}{3} d \theta\)

Integration with respect to \(\theta\)

Finally, we integrate with respect to \(\theta\): \(\int_{0}^{\pi} 64(1-\cos(\pi/6)) + \frac{14}{3} d \theta\) Since 64(1-\(\cos(\pi/6))\) and \(\frac{14}{3}\) are constants, we simply evaluate the integral as follows: \(\left(64(1-\cos(\pi/6)) + \frac{14}{3}\right) (\theta \Big|_{0}^{\pi}) = \pi\left(64(1-\cos(\pi/6)) + \frac{14}{3}\right)\). Thus, the final result is: \(\boxed{\pi\left(64(1-\cos(\pi/6)) + \frac{14}{3}\right)}\)

Key concepts

Triple Integrals

Triple integrals are a powerful tool in calculus used for calculating the volume under a surface in three-dimensional space. They extend the concept of single integrals (where you integrate over a line) and double integrals (where you integrate over a surface) into the third dimension. In spherical coordinates, the triple integral is often formatted as \(\int \int \int f(\rho, \varphi, \theta) \rho^2 \sin \varphi \, d\rho \, d\varphi \, d\theta\). Here, \(\rho\) is the radius or distance from the origin, \(\varphi\) is the polar angle from the z-axis, and \(\theta\) is the azimuthal angle in the xy-plane.
To evaluate such an integral, we perform integration in succession over each of the coordinates, typically starting with \(\rho\), then \(\varphi\), and finally \(\theta\). It allows us to find not just volumes, but also other physically meaningful quantities, such as mass, when combined with the correct density function. The exercise given is a classic example that showcases the use of spherical coordinates to simplify the evaluation process when the region has spherical symmetry.

Integration Techniques

Integration techniques are essential for solving calculus problems involving integrals, especially when dealing with complex functions or limits. In the context of spherical coordinates, integration can often involve a series of transformations and substitutions to simplify the problem.
In our given problem, we initially perform a straightforward power integration with respect to \(\rho\). This is known as using the power rule, which allows us to integrate \(\rho^2\) to get \(\frac{1}{3}\rho^3\).

• Dealing with trigonometric functions like \(\sin \varphi\) or \(\sec^3 \varphi\) often requires breaking down the integral into simpler parts.
• For instance, here \(\sin \varphi\) is handled using simple antiderivative techniques, while \(\sec^3 \varphi\) requires a bit more thought and transformation.

Early steps typically simplify the integral into familiar forms, reducing complexity and allowing for a clear path to solution. The reformatting or substitution into simpler expressions is a crucial strategy in integrating multi-variable functions.
Good practice involves recognizing the form of integral portions that are known from tables or past experiences, thus speeding up the problem-solving process.

Calculus Problems

Calculus problems involving multiple integrals can pose a significant challenge due to their complexity and the layers of abstraction they require.

• They often demand a reasonable understanding of geometry to set up the integral correctly, particularly when using non-Cartesian coordinate systems like spherical coordinates.
• A key aspect is identifying the correct limits of integration and ensuring the variables accurately represent the region of interest.

In the provided problem, we see a practical application of using these concepts to evaluate a volume integral in a spherical space.
Each integration step builds on the results of the previous, demanding precision and a systematic approach. Understanding the relationships between different coordinate systems and practicing transformations, such as from Cartesian to spherical or cylindrical, is often vital.
Additionally, problems in calculus frequently require not just computational skill but also analytical thought to interpret the results, making sense of them in a physical or geometrical context. This includes assessing whether solutions are reasonable based on the scenario presented, which ties deeply into both the setup and solving process.