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Chapter 16: Problem 53

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 \(x=a u, y=b v, z=c w.\) Find the volume of \(D.\)

Short Answer

Expert verified
Question: Determine the volume of solid \(D\), which is an ellipsoid, given the transformation \(T(u,v,w) = (au, bv, cw)\) where \(a, b, c > 0\). Answer: The volume of the ellipsoid \(D\) is \(V = \frac{4}{3} \pi abc\).

Step by step solution

01

Find the Jacobian determinant

We have been given the transformation \(T\) as follows: \(x = au, y = bv, z = cw\). To compute the Jacobian determinant, we need the partial derivatives of the transformed variables with respect to the original variables. The Jacobian matrix for the given transformation is: $$ J = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \\ \end{bmatrix} = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \\ \end{bmatrix} $$ Now, let's compute the determinant of the Jacobian matrix: $$ \begin{vmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \\ \end{vmatrix} = abc $$
02

Set up the transformed volume integral

We know that the ellipsoid equation is: $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$ We apply the given transformation: \(x = au, y = bv, z = cw\): $$ \frac{(au)^2}{a^2} + \frac{(bv)^2}{b^2} + \frac{(cw)^2}{c^2} = 1 $$ Which simplifies to: $$ u^2 + v^2 + w^2 = 1 $$ This is the equation of a unit sphere in the transformed coordinates \((u,v,w)\). To compute the volume of \(D\), we set up a triple integral as follows: $$ V = \iiint_D dV = \iiint_S abc\, du\,dv\,dw $$ Where \(S\) is the region of integration for the unit sphere in \((u,v,w)\) coordinates.
03

Compute the volume integral

The triple integral in the previous step can be transformed to spherical coordinates. Let \(u = \rho \sin{\phi} \cos{\theta}\), \(v = \rho \sin{\phi} \sin{\theta}\), and \(w = \rho \cos{\phi}\). The unit sphere has \(\rho = 1\), so we have: $$ V = abc \iiint_\Omega \rho^2 \sin{\phi}\, d\rho\,d\phi\,d\theta $$ Where \(\Omega\) is the region of integration in spherical coordinates and \(\rho\) goes from 0 to 1, \(\phi\) goes from 0 to \(\pi\), and \(\theta\) goes from 0 to \(2\pi\). $$ V = abc \int_{\rho=0}^1 \int_{\phi=0}^\pi \int_{\theta=0}^{2\pi} \rho^2 \sin{\phi}\, d\rho\,d\phi\,d\theta $$ Now, we can calculate the integral: $$ V = abc \left[ \int_{\rho=0}^1 \rho^2 d\rho \right] \left[ \int_{\phi=0}^\pi \sin{\phi} d\phi \right] \left[ \int_{\theta=0}^{2\pi} d\theta \right] $$ Calculating each individual integral: $$ V = abc \left[ \frac{\rho^3}{3} \Big|_{0}^{1} \right] \left[ -\cos{\phi} \Big|_{0}^{\pi} \right] \left[ \theta \Big|_{0}^{2\pi} \right] $$ Finally, substituting the limits: $$ V = abc \left[ \frac{1^3}{3} - \frac{0^3}{3} \right] \left[ -\cos{\pi} + \cos{0} \right] \left[ 2\pi - 0 \right] = \frac{4}{3} \pi abc $$ Thus, the volume of the ellipsoid \(D\) is \(V = \frac{4}{3} \pi abc\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Jacobian Determinant
Understanding the Jacobian determinant is crucial when evaluating integrals after a change of variables, particularly in multiple dimensions. Imagine you're working with an object that can be stretched or compressed; the Jacobian determinant essentially tells you how much the volume changes in that process.
In the case of finding the volume of an ellipsoid, we use a transformation that scales a unit sphere by factors of a, b, and c along the x, y, and z axes respectively. The Jacobian determinant of this scaling transformation, as shown in our solution, is simply the product of these scale factors: namely abc.
This value is so significant because it represents the volume scaling factor from the unit sphere to the ellipsoid when we integrate to find the volume. It's like taking a small, perfectly symmetrical object and revealing how much larger it becomes when we transform it into its final shape. Consider it a mathematical version of 'before and after' in a weight gain story; it provides concrete numbers to quantify the transformation.
Triple Integral
In calculus, when we wish to find the volume of a three-dimensional object, we naturally progress to using a triple integral. A triple integral extends the concept of an integral into three dimensions, integrating over a volume rather than an area or a line. Think of it as a way to accumulate tiny bits of volume across the entire space an object occupies.
For the ellipsoid volume problem, we set up a triple integral that sums up all the infinitesimal volume elements, dV, over the entire region D, which in this case is the transformed sphere. After applying the transformation, integrating over the volume of a sphere using cartesian coordinates would be difficult, hence we use the property that the volume element dV is the product of the Jacobian determinant and the differential elements of the transformation variables, namely the volume element of the unit sphere in u, v, w coordinates.
Spherical Coordinates
Spherical coordinates are perfect for dealing with objects that have round symmetry, like spheres or, in our case, ellipsoids. Unlike the Cartesian system which uses x, y, and z, spherical coordinates describe a point in space with three components: the radius 蟻, the polar angle 蠁, and the azimuthal angle 胃.

Transforming into Spherical Coordinates

To convert our triple integral into a form that鈥檚 more manageable, we apply spherical coordinates. This transformation aligns well with the spherical symmetry of our problem, simplifying the integration process. By performing this transformation, u, v, and w become 蟻 sin蠁 cos胃, 蟻 sin蠁 sin胃, and 蟻 cos蠁 respectively. This not only streamlines our integral but also ties neatly into the original transformation from cartesian to sphere coordinates.

Volume Element in Spherical Coordinates

One of the most beautiful aspects of spherical coordinates is the volume element, which is an elegant product of 蟻虏 sin蠁 d蟻 d蠁 d胃. This volume element ensures that we're measuring distances and angles correctly for volumes that emanate from a point, rather than those laid out in a grid, like in the Cartesian system. When integrating to find the actual volume of our ellipsoid, this volume element makes the math more tractable and intuitive.

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Most popular questions from this chapter

Find the mass of the following solids with the given density functions. Note that density is described by the function \(f\) to avoid confusion with the rudial splierical coordinate \(\boldsymbol{\rho}\). The solid cone \(\\{(r, \theta, z): 0 \leq z \leq 4,0 \leq r \leq \sqrt{3} z\) \(0 \leq \theta \leq 2 \pi\\}\) with a density \(f(r, \theta, z)=5-z\).

Suppose the density of a thin plate represented by the polar region \(R\) is \(\rho(r, \theta)\) (in units of mass per area). The mass of the plate is \(\iint_{R} \rho(r, \theta) d A .\) Find the mass of the thin half annulus \(R=\\{(r, \theta): 1 \leq r \leq 4,0 \leq \theta \leq \pi\\}\) with a density \(\rho(r, \theta)=4+r \sin \theta\).

A cylindrical soda can has a radius of \(4 \mathrm{cm}\) and a height of \(12 \mathrm{cm} .\) When the can is full of soda, the center of mass of the contents of the can is \(6 \mathrm{cm}\) above the base on the axis of the can (halfway along the axis of the can). As the can is drained, the center of mass descends for a while. However, when the can is empty (filled only with air), the center of mass is once again \(6 \mathrm{cm}\) above the base on the axis of the can. Find the depth of soda in the can for which the center of mass is at its lowest point. Neglect the mass of the can, and assume the density of the soda is \(1 \mathrm{g} / \mathrm{cm}^{3}\) and the density of air is \(0.001 \mathrm{g} / \mathrm{cm}^{3} .\)

Improper integrals Improper integrals arise in polar coordinates when the radial coordinate r becomes arbitrarily large. Under certain conditions, these integrals are treated in the usual way: $$\int_{\alpha}^{\beta} \int_{a}^{\infty} g(r, \theta) r d r d \theta=\lim _{b \rightarrow \infty} \int_{\alpha}^{\beta} \int_{a}^{b} g(r, \theta) r d r d \theta$$ Use this technique to evaluate the following integrals. $$\iint_{R} e^{-x^{2}-y^{2}} d A ; R=\left\\{(r, \theta): 0 \leq r<\infty, 0 \leq \theta \leq \frac{\pi}{2}\right\\}$$

Improper integrals Many improper double integrals may be handled using the techniques for improper integrals in one variable (Section \(8.9) .\) For example, under suitable conditions on \(f\) $$ \int_{a}^{*} \int_{\varepsilon(x)}^{h(x)} f(x, y) d y d x=\lim _{b \rightarrow \infty} \int_{a}^{b} \int_{g(x)}^{h(x)} f(x, y) d y d x $$
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