Question

Let \(D\) be the solid bounded by the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1,\) where \(a>0, b>0,\) and \(c>0\) are real numbers. Let \(T\) be the transformation \(x=\)au, \(y=b v, z=c w\) Find the average square of the distance between points of \(D\) and the origin.

Short answer

Answer: \(\frac{1}{4}(a^2 + b^2 + c^2)\)

Step-by-step solution

Set up the transformation and volume element

Apply the transformation \(x = au, y = bv, z = cw\) to the ellipsoid equation. We're given that within the ellipsoid: \(x^{2}/a^{2} + y^{2}/b^{2} + z^{2}/c^{2} = 1\) Now substitute \(x = au, y = bv, z = cw\): \((au)^2/a^2 + (bv)^2/b^2 + (cw)^2/c^2 = u^2 + v^2 + w^2 = 1\) Now we have the equation for a unit sphere in the uvw coordinate system. The volume element in the new coordinates is given by: \(dV = |\text{det}(J)| dudvdw\) where \(J\) is the Jacobian matrix and its determinant is given by: \(|\text{det}(J)| = |\text{det}(\begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix})| = abc\)

Find the Volume of the Solid

To find the volume of the solid, integrate the volume element with respect to the new coordinates over the unit sphere: \(V = \iiint_D |\text{det}(J)| dudvdw = abc \iiint_{u^2+v^2+w^2\leq1} dudvdw\) Since the integral represents the volume of the unit sphere, we know that: \(V = abc \left(\frac{4}{3}\pi\right)\)

Find the Integral of the Square of the Distance

Calculate the integral of the square of the distance, \(r^2 = x^2 + y^2 + z^2\), with respect to the volume element: \(\int_D r^2 dV = \int_D (x^2+y^2+z^2) abcdudvdw\) Using the transformation, we substitute \(x=au, y=bv, z=cw\): \(\int_D (a^2u^2 + b^2v^2 + c^2w^2) abc dudvdw\) Now, we integrate over the unit sphere in the uvw coordinate system: \(\int_{u^2+v^2+w^2\leq1} (a^2u^2 + b^2v^2 + c^2w^2) abc dudvdw\)

Calculate the Average Square of the Distance

Divide the integral from Step 3 by the total volume from Step 2 to get the average square of the distance: \(\text{Average squared distance} = \frac{\int_{u^2+v^2+w^2\leq1} (a^2u^2 + b^2v^2 + c^2w^2) abc dudvdw}{abc \left(\frac{4}{3}\pi\right)}\) \(= \frac{3}{4\pi} \int_{u^2+v^2+w^2\leq1} (a^2u^2 + b^2v^2 + c^2w^2) dudvdw\) This integral, when computed, gives: \(= \frac{3}{4\pi}\left(a^2\int_{u^2+v^2+w^2\leq1}u^2dudvdw + b^2\int_{u^2+v^2+w^2\leq1}v^2dudvdw + c^2\int_{u^2+v^2+w^2\leq1}w^2dudvdw\right)\) \(= \frac{3}{4\pi}\left(a^2 \frac{1}{3}\int_{u^2+v^2+w^2\leq1}dudvdw + b^2 \frac{1}{3}\int_{u^2+v^2+w^2\leq1}dudvdw + c^2 \frac{1}{3}\int_{u^2+v^2+w^2\leq1}dudvdw\right)\) \(= \frac{3}{4\pi}\frac{1}{3}\left(a^2 + b^2 + c^2\right)\int_{u^2+v^2+w^2\leq1}dudvdw\) \(= \frac{1}{4}(a^2 + b^2 + c^2)\) So the average square of the distance between points of solid \(D\) and the origin is: \(\frac{1}{4}(a^2 + b^2 + c^2)\)

Key concepts

Volume Element

The concept of a volume element is fundamental in the calculation of volumes for regions in three-dimensional space, especially when we are dealing with complex shapes like ellipsoids. Imagine slicing up a solid into infinitesimally small cubes or 'elements' of volume; each such 'cube' is represented by what we call the volume element. In Cartesian coordinates, this is simply represented by the product of differentials, such as dV = dx dy dz.

However, when the geometry of the region or the coordinate system changes, the volume element changes, too. For instance, in spherical or ellipsoidal coordinates, the volume element might include additional factors derived from the corresponding coordinate transformation, making it more complex than the straightforward Cartesian product.

Jacobian Determinant

When performing a coordinate transformation from one system to another, the Jacobian determinant is crucial. It tells us how the volume element changes between coordinate systems. Intuitively, you can think of it as a scaling factor that transforms the volume from one set of coordinates to the other. In the exercise, we transformed from Cartesian coordinates to uvw space, with J being the Jacobian matrix of the transformation. The determinant of this matrix gives us the scaling factor required to convert the volume element between the systems.

For our ellipsoid problem, the Jacobian determinant |det(J)| is the product of the scaling factors a, b, and c along the respective axes. This shows how a unit of volume in the uvw space corresponds to a block of volume abc in the original xyz space.

Triple Integral

Triple integrals extend the idea of single and double integrals into three dimensions, allowing us to add up values over a three-dimensional region. When calculating the volume of a solid, the triple integral sums up all the infinitesimal volume elements dV within the region of interest. This operation is vital in finding the volume of complex solids where simple geometric formulas don't apply.

In the context of this exercise, we used a triple integral to first find the volume of the unit sphere in the transformed coordinates and then to calculate the integral of the square of the distance from the origin. This demonstrates the utility of triple integrals in solving multi-dimensional problems in mathematics.

Coordinate Transformation

Coordinate transformation is a method used to simplify the integration process by changing from one set of variables to another. This technique is particularly useful when the region of interest aligns better with a different coordinate system than the standard Cartesian system. By transforming the coordinates in the original equation of the ellipsoid into a new set of variables, we get a simpler equation resembling a unit sphere.

In our case, the transformation x = au, y = bv, z = cw changed the ellipsoid equation into the equation of a sphere in uvw space. This simplified the region of integration and allowed for the utilization of the symmetry of a sphere to make the integration process more manageable. The transformative power of this technique lies in its ability to convert complicated regions into more familiar and straightforward shapes for analysis.

Solid of Revolution Volume

The volume of a solid of revolution is found by rotating a two-dimensional shape about an axis, creating a three-dimensional object. This method is often visualized by considering how a potter shapes a clay vase on a wheel. In many cases, the volume of such solids can be calculated using integration techniques.

While the ellipsoid in the given exercise is not exactly a solid of revolution, we mentioned this concept because it aligns with the idea of using integration to find a solid’s volume. Indeed, when assessing the volume of an ellipsoid or spheres - which can be viewed as a solid of revolution along multiple axes - the principles of computing revolved volumes are relevant for comprehending the general approach to determining volumes in three-dimensional space.