Problem 56 - Thomas Calculus

Question

Sphere and cylinder Find the volume of the region that lies inside the sphere \(x^{2}+y^{2}+z^{2}=2\) and outside the cylinder \(x^{2}+y^{2}=1\).

Step-by-Step Solution

Answer

The volume is \( \frac{4\pi}{3} \).

Step 1: Understand the Problem

We have a sphere defined by the equation \( x^2 + y^2 + z^2 = 2 \), and a cylinder defined by \( x^2 + y^2 = 1 \). We need to calculate the volume of the region inside the sphere but outside the cylinder.

Step 2: Visualize the Figures

The sphere is centered at the origin with radius \( \sqrt{2} \). The cylinder is a vertical infinite cylinder with radius 1, aligned along the z-axis, centered at the origin.

Step 3: Use Cylindrical Coordinates

In cylindrical coordinates, the sphere becomes \( r^2 + z^2 = 2 \) and the cylinder is \( r = 1 \). This allows easy integration by expressing the variables in terms of \( r \), \( \theta \), and \( z \).

Step 4: Set Up Integration Limits

For the region outside the cylinder, \( r \) varies from 1 to \( \sqrt{2} \). For each \( r \), \( z \) ranges from \( -\sqrt{2 - r^2} \) to \( \sqrt{2 - r^2} \). \( \theta \) will range from 0 to \( 2\pi \).

Step 5: Set Up the Volume Integral

The volume can be computed using the triple integral: \[V = \int_0^{2\pi} \int_1^{\sqrt{2}} \int_{-\sqrt{2 - r^2}}^{\sqrt{2 - r^2}} r\, dz\, dr\, d\theta\]Note the \( r \) in the integral is the Jacobian determinant for cylindrical coordinates.

Step 6: Integrate Over z

First, integrate with respect to \( z \):\[\int_{-\sqrt{2 - r^2}}^{\sqrt{2 - r^2}} r\, dz = r \times \left( 2\sqrt{2 - r^2} \right) = 2r\sqrt{2 - r^2}\]

Step 7: Integrate Over r

Now, integrate with respect to \( r \): \[\int_1^{\sqrt{2}} 2r\sqrt{2 - r^2}\, dr\]Use substitution \( u = 2 - r^2 \), hence \( du = -2r\, dr \).

Step 8: Perform Substitution

Change the limits for \( u \): when \( r = 1 \), \( u = 1 \); when \( r = \sqrt{2} \), \( u = 0 \). The integral becomes:\[-\int_1^0 \sqrt{u}\, du = \int_0^1 \sqrt{u}\, du\]

Step 9: Integrate Over u

Perform the integration:\[\int_0^1 \sqrt{u}\, du = \left[ \frac{2}{3}u^{3/2} \right]_0^1 = \frac{2}{3}\]

Step 10: Integrate Over θ

Finally, integrate over \( \theta \):\[\int_0^{2\pi} \frac{2}{3}\, d\theta = \frac{2}{3} \cdot (2\pi) = \frac{4\pi}{3}\]

Step 11: Calculate the Volume

After all the integration steps, the volume of the region inside the sphere but outside the cylinder is \( \frac{4\pi}{3} \).

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are an alternative way to express coordinates, particularly useful when dealing with problems involving symmetrical objects like cylinders. Instead of using the typical Cartesian coordinates \((x, y, z)\), cylindrical coordinates express a point in space as \((r, \theta, z)\). Here, \(r\) is the radial distance from the origin to the projection of the point in the \(xy\)-plane. The angle \(\theta\) is the angle measured counterclockwise from the positive \(x\)-axis to the line connecting the origin with the projection point in the \(xy\)-plane. The \(z\) coordinate remains the same as in Cartesian coordinates, representing height.

• For a sphere, described by \(x^2 + y^2 + z^2 = 2\) in Cartesian coordinates, it converts to \(r^2 + z^2 = 2\) in cylindrical coordinates.
• A cylinder with the equation \(x^2 + y^2 = 1\) becomes simply \(r = 1\) in these coordinates.

Switching to cylindrical coordinates can simplify the calculation, especially when the problem involves rotational symmetry around an axis.

Integration

Integration is a fundamental concept in calculus used to find areas, volumes, and other quantities under a curve or surface. Here, we employ multiple integrals to calculate the volume of the space defined in the problem. When integrating in cylindrical coordinates:

• The differential volume element is expressed as \(r \, dz \, dr \, d\theta\), rather than \(dxdydz\). This accounts for the Jacobian of the transformation from Cartesian to cylindrical, which is the \(r\) term.
• Set the limits for each variable based on the region of interest: \(r\) varies between 1 and \(\sqrt{2}\), \(z\) between \(-\sqrt{2 - r^2}\) and \(\sqrt{2 - r^2}\), and \(\theta\) spans the full circle, from 0 to \(2\pi\).

By setting these integration boundaries, the problem becomes manageable and computationally feasible. Integration is performed iteratively over \(z\), \(r\), and \(\theta\).

Triple Integral

A triple integral extends the concept of integration to three dimensions, allowing the calculation of volumes in 3D space. In this exercise, we use a triple integral to determine the volume of the region bounded by the sphere and cylinder. The process involves integrating iteratively for each of the three variables:

• Start with the innermost integral, which calculates over the variable \(z\) from \(-\sqrt{2 - r^2}\) to \(\sqrt{2 - r^2}\). This gives an output dependent on \(r\).
• Next, integrate over \(r\), utilizing a transformation if necessary, such as changing variables from \(r\) to \(u\) to simplify the computation.
• Finally, integrate over \(\theta\), ranging from 0 to \(2\pi\), to account for the rotational symmetry of the three-dimensional region.

By breaking down the volume integral into three nested integrals, we systematically solve for the volume, ultimately simplifying what could have been an overwhelming computation if attempted directly. Using a triple integral provides a structured approach to solving complex problems in calculus.