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

The value of the given triple integral is $$\frac{64\pi}{3}(1 - \frac{\sqrt{3}}{2} + 7)$$

Step-by-step solution

Change of Variables

In spherical coordinates, we have the following differential volume element: $$dV = \rho^2 \sin \varphi d\rho d\varphi d\theta$$ So, the given integral becomes: $$\int_{0}^{\pi} \int_{0}^{\pi / 6} \int_{2 \sec \varphi}^{4} (\rho^{2} \sin \varphi) dV$$

Integrate with respect to \(\rho\)

Integrating the innermost integral with respect to \(\rho\), we have: $$I_\rho = \int_{2\sec\varphi}^{4} \rho^{2} d\rho = \frac{1}{3} \rho^3 \Big|_{2\sec\varphi}^{4} = \frac{1}{3}(64 - 8^3\sec^3\varphi)$$

Integrate with respect to \(\varphi\)

Now integrating the middle integral with respect to \(\varphi\), we have: $$I_\varphi = \int_{0}^{\frac{\pi}{6}} (\frac{1}{3}(64 - 8^3\sec^3\varphi)) \sin\varphi d\varphi = \frac{1}{3} \int_{0}^{\frac{\pi}{6}} (64\sin\varphi - 8^3\sec^3\varphi \sin\varphi) d\varphi$$ To evaluate the integral with the \(\sec^3\varphi\) term, a \(u\)-substitution can be used with \(u = \sec\varphi\) and \(du = \sec\varphi\tan\varphi d\varphi\). Since \(\sin\varphi = \frac{\tan\varphi}{\sec\varphi}\), the integral can be rewritten as: $$I_\varphi = \frac{1}{3} (\int_{0}^{\frac{\pi}{6}} 64\sin\varphi d\varphi - \int_{\sec{0}}^{\sec{\frac{\pi}{6}}} 8^3 u^2 du)$$ Now, we evaluate the definite integrals separately: $$I_\varphi = \frac{1}{3} (64(-\cos\frac{\pi}{6} + \cos 0) - 8^3 (\frac{1}{3} u^3 \Big|_1^2))$$

Simplify and Integrate with respect to \(\theta\)

Simplify the expression for \(I_\varphi\) to obtain: $$I_\varphi = \frac{1}{3} (64(-\frac{\sqrt{3}}{2} + 1) - 8^3 (\frac{1}{3}(8 - 1))) = \frac{64}{3}(1 - \frac{\sqrt{3}}{2} + 7)$$ Finally, integrating with respect to \(\theta\), we have: $$I = \int_{0}^{\pi} I_\varphi d\theta = \int_{0}^{\pi} \frac{64}{3}(1 - \frac{\sqrt{3}}{2} + 7) d\theta = \frac{64}{3}(1 - \frac{\sqrt{3}}{2} + 7)\theta \Big|_0^\pi$$

Evaluate the Final Integral

Now, we evaluate the final integral: $$I = \frac{64}{3}(1 - \frac{\sqrt{3}}{2} + 7)(\pi - 0) = \frac{64\pi}{3}(1 - \frac{\sqrt{3}}{2} + 7)$$ So, the value of the given triple integral is: $$\int_{0}^{\pi} \int_{0}^{\pi / 6} \int_{2 \sec \varphi}^{4} \rho^{2} \sin \varphi d \rho d \varphi d \theta = \frac{64\pi}{3}(1 - \frac{\sqrt{3}}{2} + 7)$$

Key concepts

Triple Integrals

Triple integrals are a way of calculating the volume of a region in three-dimensional space. Imagine stacking up many thin sheets or layers to fill a volume. Each sheet represents a smaller part of the entire volume. Triple integrals help us find the total amount when these pieces are added together.

The integration occurs in three steps, each corresponding to one dimension:

• Inner integral: This is calculated first and involves integrating with respect to one variable, often denoted as \( \rho \) in spherical coordinates.
• Middle integral: After integrating the first, the result becomes an integrand for the second integral, typically with respect to \( \varphi \).
• Outer integral: Finally, we integrate the result with respect to the third variable, often \( \theta \), bringing all calculations together.

In summary, triple integrals help us compute comprehensive measures like volume in a systematic way.

Change of Variables

The change of variables is a powerful technique in integration, particularly useful in converting complex integrals into simpler ones. In the context of spherical coordinates, this transformation is instrumental because spherical geometry naturally reflects many real-world objects like planets or atoms.

When changing variables to spherical coordinates, we're replacing the traditional Cartesian coordinates (x, y, z) with \( \rho \), \( \varphi \), and \( \theta \):

• \( \rho \): Represents the radial distance from the origin to a point.
• \( \varphi \): Denotes the angle in the vertical plane from the positive z-axis.
• \( \theta \): Refers to the azimuthal angle in the xy-plane from the positive x-axis.

This method involves understanding relationships like \( x = \rho \sin\varphi \cos\theta \), which help recast the integral's limits and the volume element, \( dV = \rho^2 \sin \varphi d\rho d\varphi d\theta \). This ensures calculations align with spherical geometry, simplifying the integration process.

Integration Techniques

Integration techniques are about finding the easiest or most effective way to solve an integral. In solving the provided problem, various techniques illustrate different strategies to overcome challenges in calculation.

The primary technique used involves the method of integration by substitution. This is evident when approaching the integral with a \( \sec^3\varphi \) term. Here, the substitution \( u = \sec\varphi \), transforms difficult-to-handle trigonometric expressions into simpler algebraic forms.

Steps for substitution involve:

• Choosing a substitution that will simplify the expression.
• Calculating the differential \( du \) to replace the variable and its differential in the integral.
• Resolving the limits of integration to the new variable \( u \).

These techniques not only aid in simplifying the problem but also open paths to calculating integrals that might otherwise appear unsolvable. Mastery of these techniques is essential for tackling complex integrals in higher mathematics.