Problem 19 - Thomas Calculus

Question

Evaluate \begin{equation} \begin{array}{c} {\iiint|x y z| d x d y d z} \\ {\text { over the solid ellipsoid }} \\ {\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} \leq 1} \\ {\text { (Hint: Let } x=a u, y=b v, \text { and } z=c w . \text { Then integrate over an }} \\ {\text { appropriate region in } u v w \text { -space.) }}\end{array} \end{equation}

Step-by-Step Solution

Answer

The integral evaluates to 0 due to symmetry.

Step 1: Change of Variables

We start by making the change of variables suggested by the hint. Let \( x = a u, y = b v, z = c w \). This means that \( u = \frac{x}{a}, v = \frac{y}{b}, w = \frac{z}{c} \), transforming the ellipsoid into the unit sphere given by \( u^2 + v^2 + w^2 \leq 1 \).

Step 2: Determine the Jacobian

Next, calculate the determinant of the Jacobian matrix for the transformation. The Jacobian is given by \( \frac{\partial(x, y, z)}{\partial(u, v, w)} = \begin{vmatrix}a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c \end{vmatrix} = abc \).

Step 3: Set up the Integral with New Variables

The integral now becomes \[ \iiint_{u^2 + v^2 + w^2 \leq 1} |a b c u v w| \, abc \, du \, dv \, dw \]. This simplifies to \[ (abc)^2 \iiint_{u^2 + v^2 + w^2 \leq 1} |u v w| \, du \, dv \, dw \].

Step 4: Simplify the Integral Limits

We are integrating over a sphere of radius 1 in \( (u, v, w) \)-space, which is a well-known symmetry, allowing us to use spherical coordinates or symmetry arguments for simplification.

Step 5: Evaluate the Integral

Using spherical symmetry, note that the integral of an odd function, like \(|u v w|\), over a perfectly symmetric \region like the sphere results in 0 because the negative parts cancel out the positive parts.

Key Concepts

Ellipsoid

An ellipsoid is a three-dimensional geometric shape that resembles a distorted sphere. Unlike a perfect sphere, an ellipsoid has axes of different lengths. The primary formula for an ellipsoid is given by:

• \(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1\)

Where \(a\), \(b\), and \(c\) represent the semi-axis lengths along the x, y, and z directions respectively. This expression implies that the ellipsoid is contained within a unit sphere shape when considering the scaled variables.

Ellipsoids are important in many fields such as physics, engineering, and graphics, as they model various physical phenomena and structures. They can be visualized as the 3D analog of an ellipse and help in simplifying complex volume integrals, like in this exercise.

Change of Variables

In calculus, the change of variables technique is a powerful tool that simplifies integration processes, especially in multiple dimensions. When dealing with complex regions like an ellipsoid, directly integrating can be quite difficult.

• To convert the cumbersome shape of an ellipsoid into a simpler form like a sphere, new variables can be introduced.
• In our problem, this is achieved by setting: \(x = au\), \(y = bv\), \(z = cw\).

This transformation is essentially scaling the axes of the ellipsoid to transform the inequality constraint of the ellipsoid into a unit sphere condition \(u^2 + v^2 + w^2 \leq 1\).

Such transformations not only simplify the region over which the integral is evaluated but also the function itself, leveraging symmetries and reducing complexity. Thus, a change of variables is an excellent strategy for tackling integrals over non-standard regions.

Jacobian

The Jacobian is a determinant associated with the change of variables in multivariable integration. It accounts for how much the infinitesimal volume element changes under the transformation. This is critical when integrating after a change of variables, ensuring that the integral correctly adjusts for the new coordinate system.

• For the transformation \((x = au, y = bv, z = cw)\), the Jacobian is calculated as \(abc\).
• This comes from the determinant of the matrix formed by partial derivatives: \(\begin{vmatrix}a & 0 & 0 \0 & b & 0 \0 & 0 & c\end{vmatrix}\).

This result indicates that every small volume in the \((u,v,w)\)-space relates to a volume in \((x,y,z)\)-space scaled by \(abc\).

Thus, integrating in the new variable space requires multiplying the integrand by the Jacobian determinant. It is vital for accurately transforming both the integrand and the domain of integration.

Spherical Coordinates

Spherical coordinates provide a way to express points in three-dimensional space with radii and angles, which is useful when dealing with shapes and regions exhibiting symmetry, like the sphere obtained after the change of variables.

• In spherical coordinates, any point is described by \((\rho, \theta, \phi)\), where \(\rho\) is the radius, \(\theta\) the azimuthal angle, and \(\phi\) the polar angle.
• This system is advantageous because it simplifies the integration over spherical regions by leveraging radial symmetry.

For our problem, considering spherical coordinates makes evaluating the integral straightforward, especially when the sphere's symmetry significantly cancels out complex parts of calculations.

As a result, even complicated integrals, such as triple integrals over a sphere, become much more manageable. The power of spherical coordinates lies in how they elegantly handle multidimensional angles and distances, wrapping complexities into more digestible terms.