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, $y=bv,z=cw$ Evaluate ${\iiint }_D|xyz|dA$

Short answer

Short Answer: After changing the variables of integration using the given transformation - $x=au$, $y=bv$, and $z=cw$, we find that the transformed integrand is given by $|abcuvw|$. We calculate the Jacobian of the transformation and find that it is equal to abc. Then we evaluate the triple integral over the unit ball in the new uvw-coordinate system. Since the function is odd with respect to u, v, and w and the integration domain is symmetric, the final integral evaluates to 0. Thus, the volume of the solid bounded by the given ellipsoid is 0.

Step-by-step solution

Determine the new expression for the integrand under the transformation:

We are given a transformation T with $x=au$, $y=bv$, and $z=cw$. Using these expressions, we rewrite the integrand $xyz$: Integrand: $|xyz|=|(au)(bv)(cw)|=|abcuvw|$.

Find the Jacobian of the transformation:

The Jacobian J is the determinant of the matrix of partial derivatives of the transformation with respect to the new variables. In our case, we have: J = $\left|\begin{array}{ccc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}& \frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}& \frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}& \frac{\partial z}{\partial w}\end{array}\right|=\left|\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right|=abc$.

Substitute the transformation and Jacobian into the integral:

We have: ${\iiint }_D|xyz|dA={\iiint }_{T(D)}|abcuvw|abcdudvdw$.

Determine the transformed domain T(D):

The equation of the ellipsoid is given by: $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$. Substitute the transformation $x=au$, $y=bv$, and $z=cw$: $\frac{(au{)}^2}{a^2}+\frac{(bv{)}^2}{b^2}+\frac{(cw{)}^2}{c^2}=1⟹u^2+v^2+w^2=1$. Thus, the transformed domain T(D) is the unit ball in the uvw-coordinate system.

Evaluate the integral:

We now evaluate the integral over the unit ball: ${\iiint }_{T(D)}|abcuvw|abcdudvdw=(abc{)}^2{\iiint }_{u^2+v^2+w^2\le 1}uvwdudvdw$. We can see that the function is odd with respect to u, v, and w. The integration domain is symmetric with respect to the u, v, and w planes. Thus, the integral is equal to 0: $(abc{)}^2{\iiint }_{u^2+v^2+w^2\le 1}uvwdudvdw=0$. So, the final answer is: ${\iiint }_D|xyz|dA=0$.

Key concepts

Ellipsoid Volume

The concept of an ellipsoid is crucial in understanding the geometric context of many integration problems. In layperson's terms, an ellipsoid is a three-dimensional figure that looks like a stretched or squashed sphere. The volume of a regular ellipsoid is given by the formula $\frac{4}{3}\pi abc$, where $a$, $b$, and $c$ are the lengths of the semi-axes along the x, y, and z dimensions respectively.

In calculus, when we talk about the volume of an ellipsoid, we often mean the volume bounded by the surface of the ellipsoid. Think of it like the amount of water such an ellipsoid-shaped container could theoretically hold. When dealing with triple integrals, calculators can essentially 'add up' infinitesimally small volumes within the confines of this three-dimensional shape to find the total volume.

Coordinate Transformation

Coordinate transformation is like changing the language we use to describe locations in space. Most students are familiar with Cartesian coordinates (x, y, z), which is like giving an address using a city grid. However, certain shapes are more naturally described using different coordinate systems, such as cylindrical or spherical coordinates.

For the ellipsoid volume problem, we use a simple scaling transformation, where each Cartesian coordinate is multiplied by a different constant (a, b, c for x, y, z respectively). This transformation maps the ellipsoid to a unit sphere, making our integration problem simpler and symmetrical. Visualize this as taking an oddly shaped balloon and deflating it into a perfect, easy-to-measure sphere.

Jacobian Determinant

The Jacobian determinant is a little like the adjustment factor when changing between units of measurement, for instance, from inches to centimeters. In the context of integration, it tells us how much the volume element (the tiny 'box' of space we're adding up) changes size when we switch coordinate systems.

It is calculated by taking the determinant of a matrix made from the partial derivatives of the transformed coordinates with respect to the original coordinates. In our ellipsoid problem, the matrix is diagonal with entries (a, b, c), and so the Jacobian determinant is simply the product abc. Just as you'd multiply by 2.54 to convert inches to centimeters, we multiply our integral by the Jacobian determinant to correctly measure volume in the transformed coordinates.

Multiple Integration

Multiple integration refers to performing integration over more than one dimension. For a three-dimensional shape, we use triple integration. You can visualize triple integration as building up the volume of a 3D shape slice by slice, with every infinitesimal slice having its own infinitesimal area.

When we work in a transformed coordinate system that matches the symmetry of the problem, like mapping an ellipsoid to a unit sphere, it allows us to easily express and integrate over this space. In our case, we could break up the unit sphere (the transformed ellipsoid) into an infinite number of infinitesimally thin slices from its center to its surface, simplifying the calculation of the original ellipsoidal volume.