Calculus III Advanced (Course) (11.7) (Homework)

Section 11.7 Homework

From Calculus 10e by Larson and Edwards, p. 809. Exercises 4, 8, 26, 32, 38, 44

Exercise 11.7.4 Cylindrical-to-Cartesian Conversion

Convert

$$(x,y,z)=\left( 6,-\frac{\pi}{4}, 2 \right ) $$

from cylindrical to cartesian.
Solution

$$x=6 \cos \left ( -\frac{\pi}{4} \right) = 3\sqrt{2} $$

$$y=6 \sin \left ( -\frac{\pi}{4} \right) = -3\sqrt{2}$$

\(z=2\)

the cartesian point is \((3\sqrt{2},-3\sqrt{2},2)\).

Exercise 11.7.8 Cartesian-to-Cylindrical Conversion

Convert the point

\((x,y,z)=(2\sqrt{2},-2\sqrt{2},4)\)

from cartesian to cylindrical.
Solution

\(r^{2}=(2\sqrt{2})^{2} + (-2\sqrt{2})^{2} \rightarrow r=\pm 4\)

$$\tan \theta = \frac{-2\sqrt{2}}{2\sqrt{2}}= -1 \rightarrow \theta = \arctan(-1)+n\pi= -\frac{\pi}{4}+n\pi$$

\(z=4\)

There are two possible answers, \(r >0\) and \(n=0\)

$$\left ( 4, -\frac{\pi}{4}, 4 \right ) $$

and

$$\left ( -4, -\frac{5\pi}{4}, 4 \right ) $$

Exercise 11.7.26 Cylindrical-to-Cartesian Equation Conversion

Convert the equation

\(z=r^{2}\cos^{2} \theta\)

from cylindrical to cartesian and sketch the graph.
Solution

\(r^{2}\cos^{2} \theta\)	\(=z \)	Cylindrical Equation
\(r^{2}\cos^{2} \theta -z\)	\(= 0 \)	Subtract \(z\) from both sides
\(r^{2} ( 1/2(1+\cos(2 \theta))) -z\)	\(= 0 \)	Trigonometric Identity
\(r^{2} ( 1/2+1/2\cos(2 \theta)) -z\)	\(= 0 \)
\(1/2r^{2}+1/2r^{2}\cos(2 \theta) -z\)	\(= 0 \)
\(1/2r^{2}+1/2r^{2}(\cos 2\theta - \sin^{2} \theta) -z\)	\(= 0 \)	Trigonometric Identity
\(1/2r^{2}+1/2r^{2}\cos 2\theta - 1/2r^{2}\sin^{2} \theta -z\)	\(= 0 \)
\(1/2(x^{2} + y^{2})+x^{2} - y^{2} -z\)	\(=0\) Replace \(1/2r \cos \theta\) with \(x\) and \(1/2r \sin \theta\) with \(y\)

Exercise 11.7.32 Cartesian-to-Spherical Point Conversion

Convert the point \((x,y,z)=(2,2,4\sqrt{2})\) from cartesian to spherical.
Solution

\(r=\sqrt{2^{2}+2^{2}} = \sqrt{8}\)

\(\tan \theta = 2/2 = 1 \rightarrow \theta = \pi/4 + \pi n \)

\(z=4\sqrt{2}\)

Exercise 11.7.38 Spherical-to-Cartesian Point Conversion

Convert the spherical point \(( \rho,\theta,\phi)=(9,\pi/4,\pi)\) to cartesian.
Solution

\(x=9 \sin \pi \cos \pi/4 = 9 \cdot 0 \sqrt{2}/2 = 0\)

\(y=9 \sin \pi \sin \pi/4 = 9 \cdot 0 \sqrt{2}/2 = 0 \)

\(z=9 \cos \pi = -9 \)

Exercise 11.7.44 Cartesian-to-Spherical Equation Conversion

Convert the cartesian equation \(x^{2}+y^{2}-3z^{2} = 0\) to a spherical equation.
Solutions
Substitute cartesian for spherical points.

\(x^{2}+y^{2}-3z^{2} \)	\(= 0 \)
\( \rho^{2} \sin^{2} \theta \cos^{2} \theta + \rho^{2} \sin^{2} \phi \sin^{2} \theta -3(\rho^{2} \cos^{2} \phi)\)	\(= 0 \)
\( \rho^{2} \sin^{2} \phi ( \cos^{2} \theta + \sin^{2} \theta) -3(\rho^{2} \cos^{2} \phi)\)	\(= 0 \)
\( \rho^{2} \sin^{2} \phi -3(\rho^{2} \cos^{2} \phi)\)	\(= 0 \)
\( \rho^{2} \sin^{2} \phi \)	\(= 3(\rho^{2} \cos^{2} \phi) \)
$$ \frac{\sin^{2} \phi}{\cos^{2} \phi} $$	\(=3\)
\( \tan^{2} \phi \)	\(= 3 \)
\( \tan \phi \)	\(= \sqrt{3} \)
\( \phi \)	\(= \pi/3 + \pi n \)

This produces \( \phi \) with two possible values

$$ \phi = \frac{\pi}{3} \text{ or } \frac{4\pi}{3} $$

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Parent Article: Calculus III Advanced (Course)