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Chapter 13: Problem 39

Evaluate the Jacobians \(J(u, v, w)\) for the following transformations. $$x=v w, y=u w, z=u^{2}-v^{2}$$

Short Answer

Expert verified
Question: Determine the Jacobian, J(u, v, w), given the transformations below: x(u, v, w) = vw y(u, v, w) = uw z(u, v, w) = u^2 - v^2 Answer: J(u, v, w) = 4uvw

Step by step solution

01

Calculate Partial Derivatives

Compute the partial derivatives of the given transformations: \(\frac{\partial x}{\partial u}, \frac{\partial x}{\partial v}, \frac{\partial x}{\partial w}, \frac{\partial y}{\partial u}, \frac{\partial y}{\partial v}, \frac{\partial y}{\partial w}, \frac{\partial z}{\partial u}, \frac{\partial z}{\partial v}, \frac{\partial z}{\partial w}\). $$\frac{\partial x}{\partial u} = 0$$ $$\frac{\partial x}{\partial v} = w$$ $$\frac{\partial x}{\partial w} = v$$ $$\frac{\partial y}{\partial u} = w$$ $$\frac{\partial y}{\partial v} = 0$$ $$\frac{\partial y}{\partial w} = u$$ $$\frac{\partial z}{\partial u} = 2u$$ $$\frac{\partial z}{\partial v} = -2v$$ $$\frac{\partial z}{\partial w} = 0$$
02

Write Out the Jacobian Matrix

Write the 3x3 Jacobian Matrix using the calculated partial derivatives: $$ J(u, v, w) = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{bmatrix} = \begin{bmatrix} 0 & w & v\\ w & 0 & u\\ 2u & -2v &0 \end{bmatrix} $$
03

Find Determinant of Jacobian

Calculate the determinant of the Jacobian Matrix to obtain J(u, v, w): $$ J(u, v, w) = \begin{vmatrix} 0 & w & v\\ w & 0 & u\\ 2u & -2v &0 \end{vmatrix} =(0)\cdot((0)(0)-(-2v)(u))-w\cdot(w(0)-(u)(-2v)) + v\cdot(w(-2v)-2u(0)) = 2uvw - (-2uvw) = 4uvw $$ Thus, the Jacobian is \(J(u, v, w) = 4uvw\).

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Most popular questions from this chapter

Let \(R\) be the region bounded by the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1,\) where \(a>0\) and \(b>0\) are real numbers. Let \(T\) be the transformation \(x=a u, y=b v\) Find the center of mass of the upper half of \(R(y \geq 0)\) assuming it has a constant density.

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, \(y=b v, z=c w\) Evaluate \(\iiint_{D}|x y z| d A\)

A thin (one-dimensional) wire of constant density is bent into the shape of a semicircle of radius \(a\). Find the location of its center of mass. (Hint: Treat the wire as a thin halfannulus with width \(\Delta a,\) and then let \(\Delta a \rightarrow 0\).)

Evaluate the following integrals in spherical coordinates. $$\int_{0}^{2 \pi} \int_{0}^{\pi / 3} \int_{0}^{4 \sec \varphi} \rho^{2} \sin \varphi d \rho d \varphi d \theta$$

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 \(0.001 \mathrm{g} / \mathrm{cm}^{3}\)
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