Problem 77 - Thomas Calculus

Question

a. Express the area \(A\) of the cross-section cut from the ellipsoid $$ x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{9}=1 $$ by the plane \(z=c\) as a function of \(c .\) (The area of an ellipse with semiaxes \(a\) and \(b\) is \(\pi a b . )\) b. Use slices perpendicular to the \(z\) -axis to find the volume of the ellipsoid in part (a). c. Now find the volume of the ellipsoid $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 $$ Does your formula give the volume of a sphere of radius \(a\) if \(a=b=c ?\)

Step-by-Step Solution

Answer

a) \(A(c) = 2\pi \left(1 - \frac{c^2}{9}\right)\); b) Volume = \(12\pi\); c) Ellipsoid volume is \(\frac{4}{3}\pi abc\), matches sphere when \(a=b=c=r\).

Step 1: Find the Equation of the Cross-Section

Given the ellipsoid equation, substitute \(z = c\) into \(x^2 + \frac{y^2}{4} + \frac{z^2}{9} = 1\). This becomes \(x^2 + \frac{y^2}{4} + \frac{c^2}{9} = 1\). Simplify to find the equation of the cross-sectional ellipse: \(x^2 + \frac{y^2}{4} = 1 - \frac{c^2}{9}\).

Step 2: Express Semi-Axes of the Cross-Section

From the simplified equation \(x^2 + \frac{y^2}{4} = 1 - \frac{c^2}{9}\), identify the semi-axes of the ellipse. The semi-major axis is \(a = \sqrt{1 - \frac{c^2}{9}}\) for \(x\) and \(b = 2\sqrt{1 - \frac{c^2}{9}}\) for \(y\).

Step 3: Area of the Elliptical Cross-Section

Use the area formula for an ellipse \(A = \pi a b\). Substitute the semi-axes \(a = \sqrt{1 - \frac{c^2}{9}}\) and \(b = 2\sqrt{1 - \frac{c^2}{9}}\) into the formula. Therefore, the area \(A(c)\) is given by: \(A(c) = \pi \left(\sqrt{1 - \frac{c^2}{9}}\right) \left(2\sqrt{1 - \frac{c^2}{9}}\right) = 2\pi \left(1 - \frac{c^2}{9}\right)\).

Step 4: Volume using Disks Perpendicular to the z-axis

Take slices perpendicular to the \(z\)-axis, each with thickness \(dz\). The volume \(V\) of a disk at height \(z\) is \(A(z) \, dz\). Integrate from \(-3\) to \(3\) (the bounds of \(z\)) for the full volume: \(V = \int_{-3}^{3} 2\pi \left(1 - \frac{z^2}{9}\right) \, dz\).

Step 5: Evaluate the Integral for Volume

Calculate \(V = 2\pi \int_{-3}^{3} \left(1 - \frac{z^2}{9}\right) dz\). Simplify to \(V = 2\pi \left[ z - \frac{z^3}{27} \right]_{-3}^{3}\). This evaluates to \(V = 2\pi \left[ (3 - \frac{27}{27}) - (-3 + \frac{-27}{27}) \right]\), resulting in \(V = 2\pi \times 6 = 12\pi\).

Step 6: General Volume Formula for Ellipsoid

For the ellipsoid \(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\), the volume formula is \(V = \frac{4}{3}\pi abc\).

Step 7: Check the Volume for a Sphere

If \(a = b = c = r\), substitute into the ellipsoid volume formula to get \(V = \frac{4}{3}\pi r^3\). This confirms that the volume formula reduces correctly to the volume of a sphere with radius \(r\).

Key Concepts

Cross-Section Area

When an ellipsoid is intersected by a plane, the resulting shape is an ellipse. This intersection is called a cross-section. To find the area of this elliptical cross-section, we need to first determine the equation of the cross-section by substituting a constant value for one of the variables, in this case, the height or axis variable.
For the ellipsoid described by the equation \[ x^2 + \frac{y^2}{4} + \frac{z^2}{9} = 1 \]when sliced by the plane at height \(z=c\), we substitute to get:\[ x^2 + \frac{y^2}{4} = 1 - \frac{c^2}{9} \]
This equation shows us an ellipse with semi-axes along x and y.

• The semi-major axis \(a\) of the ellipse is \(\sqrt{1 - \frac{c^2}{9}}\) along the x-axis.
• The semi-minor axis \(b\) of the ellipse is \(2\sqrt{1 - \frac{c^2}{9}}\) along the y-axis.

To find the area of the ellipse, we use the formula for the area of an ellipse, \(A = \pi a b\). Substituting the semi-axes values, we find the cross-sectional area:\[ A(c) = 2\pi \left(1 - \frac{c^2}{9}\right) \]This gives insight into how the cross-section area varies with \(c\), the position of the plane.

Elliptical Geometry

Elliptical geometry involves understanding shapes like ellipses, which are formed by slicing a cone or an ellipsoid at different angles. An ellipse has two main axes: the semi-major axis and the semi-minor axis.
The lengths of these axes determine the shape and area of the ellipse. In the context of the exercise, we see:

• The semi-major axis \(a\) is related to the variable along which the ellipse is longest.
• The semi-minor axis \(b\) is linked to the direction in which the ellipse is shorter.

For the given ellipsoid, which is symmetric around the origin, the geometry involves understanding how these semi-axes change when intersected by a plane.
For understanding this, symmetry and stretching or squeezing factors play a key role:

• The stretching factor associated with the y-axis is influenced by its division by 4 in the ellipsoid equation, giving a squeezed effect to the axis associated with \(y^2 \).
• The z-axis is divided by 9, which elongates the ellipsoid along z, affecting how slices perpendicular to z manifest as ellipses.

Elliptical geometry principles help you determine these semi-axes modifications and thus the derived area and volume calculations.

Integration Techniques

Calculating the volume of an ellipsoid using integral calculus involves slicing the shape into infinitesimally thin disks or layers. Each slice has its own area which changes as you move along the height.
To determine the integral necessary for volume:

• Slices are taken perpendicular to the z-axis, where each slice at a height \(z\) can be viewed as a disk with area \(A(z)\).
• Each disk's volume is calculated over an infinitesimally small thickness \(dz\).

The integral to find the volume \(V\) sums over these slices:\[ V = \int_{-3}^{3} 2\pi \left(1 - \frac{z^2}{9}\right) dz \]
This integral evaluates the contribution of each slice from the bottom of the ellipsoid to the top. Calculus concepts such as definite integrals and substitution can help simplify this step.

• Using symmetry, often calculations are only needed for half of the object, with results doubled for symmetry about an axis.
• The formula ultimately leading to the ellipsoid's volume\[ V = \frac{4}{3}\pi abc \] exemplifies integrating functions of the type seen in ellipses and higher-dimensional shapes.

Using integration, we derive volumes for both general shapes like our ellipsoid and special cases such as spheres.