Question

Use spherical coordinates to find the volume of the following solids. The solid bounded by the sphere \(\rho=2 \cos \varphi\) and the hemisphere \(\rho=1, z \geq 0\)

Short answer

Answer: The volume of the solid is \(\frac{4\pi}{3}\).

Step-by-step solution

Find the limits of integration for each coordinate

First, we have to find the values of \(\rho\), \(\varphi\), and \(\theta\) corresponding to the boundaries of the solid. For \(\rho\) values, we know that the solid is bounded by two surfaces: the sphere \(\rho = 2\cos\varphi\) and the hemisphere \(\rho = 1\). Thus, \(\rho\) varies from \(0\) to \(2\cos\varphi\) inside the solid. To find the range for \(\varphi\), consider the sphere \(\rho = 2\cos\varphi\). When \(\rho = 0\) (at the origin), we have \(\varphi = \frac{\pi}{2}\) (the equator). When the sphere touches the hemisphere of \(\rho = 1\), we have \(\varphi = 0\). Therefore, \(\varphi\) varies from \(0\) to \(\frac{\pi}{2}\). The \(z\)-axis is at \(\theta=0\). There are no boundaries specified along the \(\theta\) direction. Therefore, since the solid is symmetrical about this axis, \(\theta\) must vary from 0 to \(2\pi\).

Integrate the volume differential element

We now integrate the volume differential element, \(dV = \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta\), over the solid region: $$ V = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\cos\varphi} \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta$$

Evaluate the integral

We can evaluate the integral by integrating with respect to each coordinate, one at a time: $$V = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \left[\frac{\rho^3}{3}\sin\varphi\right]_0^{2\cos\varphi} d\varphi \, d\theta = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \frac{8}{3}\cos^3\varphi\sin\varphi \, d\varphi \, d\theta$$ $$V = \int_{0}^{2\pi} \left[\frac{8}{3}\left(-\frac{1}{4}\cos^4\varphi\right)\right]_0^{\frac{\pi}{2}} d\theta= \int_{0}^{2\pi} -\frac{2}{3} d\theta$$ Finally: $$V = \left[-\frac{2}{3}\theta\right]_{0}^{2\pi} = -\frac{4\pi}{3}$$ However, the volume should always be positive, so: $$V = \frac{4\pi}{3}$$

Final Answer

The volume of the solid bounded by the sphere \(\rho=2 \cos \varphi\) and the hemisphere \(\rho=1, z \geq 0\) is: $$V = \frac{4\pi}{3}$$

Key concepts

Triple Integration

In the context of finding volumes of solids, triple integration is a versatile technique used to sum up an infinite number of infinitesimally small volume elements over a three-dimensional region. The idea is to split the solid into tiny parts, find the volume of each part, and then add them all together.

When using triple integration for finding volumes, you should think of it as a three-step process. In the first step, you allignate the proper coordinate system (Cartesian, cylindrical, or spherical) based on the symmetry of the solid. Then, decide the limits of integration that correctly represent the boundaries of the solid in question. Lastly, execute the integration process over the three variables, one by one, which in our case, are the spherical coordinates: \( \rho \) (radius), \( \varphi \) (polar angle), and \( \theta \) (azimuthal angle). This step-by-step approach ensures each dimension is appropriately accounted for to compute the total volume.

Solid of Revolution

A solid of revolution is a solid created by rotating a two-dimensional shape around an axis. The resulting three-dimensional object has a symmetry that can often simplify the process of finding its volume. Spherical coordinates are particularly helpful when dealing with solids of revolution because they are designed to describe points in 3D space based on their distance from the origin and two angles, capturing the essence of rotational symmetry.

For example, when a semi-circle in the \( xz \) plane is revolved around the \( z \) axis, we get a hemisphere. In spherical coordinates, this hemisphere can be described simply by fixing one of the limits of \( \rho \) at a constant value, such as \( \rho=1 \) in the exercise provided. Their radial symmetry allows the use of \( \varphi \) and \( \theta \) to determine the extent of the volume without complicated bounds typically found in Cartesian coordinates.

Volume Differential Element

The volume differential element in spherical coordinates, \( dV = \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta \), encapsulates the essence of triple integration in spherical coordinates. It represents an infinitesimally small 'chunk' of volume within a sphere, influenced by the radius \( \rho \) and the angles \( \varphi \) and \( \theta \).

In our exercise, the differential element expands and contracts with \( \rho \) and sweeps through the various angles to cover the entire volume. Introducing \( \rho^2 \sin\varphi \) accounts for the 'slicing' and 'stacking' needed to fill the volume completely. As \( \varphi \) goes from 0 to \( \frac{\pi}{2} \) (or from the positive \( z \) axis to the xy plane), and as \( \theta \) goes from 0 to \( 2\pi \) (a full rotation around the z-axis), the differential element sweeps out the entire region of interest. By integrating this element over its limits, we sum all such infinitesimal volumes to find the total volume of the solid.