Question

Use spherical coordinates to find the volume of the following solids. The solid bounded by the cylinders \(r=1\) and \(r=2,\) and the cones \(\varphi=\pi / 6\) and \(\varphi=\pi / 3\)

Short answer

Answer: The volume of the solid is (7π/3) cubic units.

Step-by-step solution

Setting up the integral

To find the volume of the solid, we will integrate the volume element in spherical coordinates over the given region. The volume element in spherical coordinates is given by \(dV = r^2 \sin(\varphi) dr d\varphi d\theta\). The limits of integration will be determined by the radius and angle bounds given by the cylinders and cones.

Determining the limits of integration

The radius will range between the two cylindrical bounds, so \(r\) will go from 1 to 2. The angle \(\varphi\) will be bounded by the two cones, so \(\varphi\) will range from \(\pi/6\) to \(\pi/3\). Finally, the angle \(\theta\) will vary between 0 and \(2\pi\) since the solid spans the entire circle in the \(xy\)-plane. Thus, the limits of integration are: - \(r\): 1 to 2 - \(\varphi\): \(\pi/6\) to \(\pi/3\) - \(\theta\): 0 to \(2\pi\)

Setting up and evaluating the triple integral

Now we can set up the triple integral for the volume: $$V = \int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \int_{1}^{2} r^2 \sin(\varphi) dr d\varphi d\theta$$ First, we will integrate with respect to \(r\): $$V = \int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \left[\frac{1}{3}r^3\right]_{1}^{2} \sin(\varphi) d\varphi d\theta$$ Now, we evaluate the expression inside the brackets: $$V = \int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \left( \frac{8}{3} - \frac{1}{3} \right) \sin(\varphi) d\varphi d\theta = \int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \frac{7}{3} \sin(\varphi) d\varphi d\theta$$ Next, we will integrate with respect to \(\varphi\): $$V = \int_{0}^{2\pi} \left[-\frac{7}{3}\cos(\varphi)\right]_{\pi/6}^{\pi/3} d\theta$$ Now, we evaluate the expression inside the brackets: $$V = \int_{0}^{2\pi} \left( -\frac{7}{6} + \frac{7}{3} \right) d\theta = \int_{0}^{2\pi} \frac{7}{6} d\theta$$ Finally, we will integrate with respect to \(\theta\): $$V = \left[\frac{7}{6}\theta\right]_{0}^{2\pi}$$ Now, we evaluate the expression inside the brackets: $$V = \frac{7}{6}(2\pi) - \frac{7}{6}(0) = \frac{7\pi}{3}$$ Thus, the volume of the solid is \(\frac{7\pi}{3}\) cubic units.

Key concepts

Volume of Solids

Finding the volume of solids using spherical coordinates is a fascinating way of handling three-dimensional space problems. In our particular problem, we have a solid bounded by two cylinders and two cones.
This means the solid is essentially a 'slice' of a spherical shell. When attempting to compute the volume of such a solid, we consider a small element of volume and integrate over the entire region. In spherical coordinates, the volume element is denoted as \(dV = r^2 \sin(\varphi) \, dr \, d\varphi \, d\theta\), where \(r\) is the radius, \(\varphi\) is the angle from the positive z-axis (similar to latitude), and \(\theta\) is the angle in the \(xy\)-plane (similar to longitude).
Understanding how to transform solids bounded by straightforward geometric shapes (like cylinders and cones) into spherical coordinates is key for analyzing and calculating their volumes.

Triple Integral

A triple integral is a powerful tool for calculating volumes and other quantities extended across a three-dimensional space. In our exercise, we used a triple integral to compute the total volume of the solid.
The expression \( \int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \int_{1}^{2} r^2 \sin(\varphi) \, dr \, d\varphi \, d\theta \) represents the sum of infinitely many small volume elements across the specified range. To solve it, we carefully dissect it by integrating one variable at a time:

• Start with \(r\), the radial distance, capturing the volume over the inner and outer cylinder radii.
• Next, handle \(\varphi\), the spherical polar angle that captures the space between the cones.
• Finally, integrate \(\theta\), the azimuthal angle representing a full circle around the z-axis.

This systematic approach ensures each element contributes accurately to the total volume.

Limits of Integration

Determining the correct limits of integration is crucial for accurately setting up the triple integral. In the context of spherical coordinates, the limits correspond to physical constraints set by the solid's geometric boundaries. For the given problem:

• The radial distance \(r\) is bounded between the two cylinders at \(r=1\) and \(r=2\), hence \(r\) varies from 1 to 2.


• The angle \(\varphi\), governed by the cones, stretches from \(\pi/6\) to \(\pi/3\). This range represents the space the solid occupies relative to the z-axis.


• Lastly, the azimuthal angle \(\theta\) naturally spans a full circle in the \(xy\)-plane, from 0 to \(2\pi\).

Choosing these limits correctly ensures the integration accounts for exactly the volume within the solid's boundaries. Interpreting these constraints in spherical terms makes it more intuitive to visualize and solve the problem.