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Calculus 3 : 3-Dimensional Space

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Example Questions

Example Question #511 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (83,88^{\circ},54^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(-46.35,-1.64,-68.83)

(-68.79,-2.44,-46.38)

(1.7,48.76,67.15)

(2.34,67.11,48.79)

Correct answer:

(2.34,67.11,48.79)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=83sin(54^{\circ})cos(88^{\circ})=2.34\\&y=83sin(54^{\circ})sin(88^{\circ})=67.11\\&z=83cos(54^{\circ})48.79\end{align*}$

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Example Question #512 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (29,-114^{\circ},123^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(-9.89,-22.22,-15.79)

(-15.95,20.21,-13.34)

(6.42,14.43,24.32)

(-8.26,10.47,-25.75)

Correct answer:

(-9.89,-22.22,-15.79)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=29sin(123^{\circ})cos(-114^{\circ})=-9.89\\&y=29sin(123^{\circ})sin(-114^{\circ})=-22.22\\&z=29cos(123^{\circ})-15.79\end{align*}$

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Example Question #513 : 3 Dimensional Space

Convert the following vector from Cartesian coordinates into spherical coordinates:

$(x^{-1}y, xz, 3y^2)$

Possible Answers:

$(\tan \phi, r^2 \sin\theta\cos^2 \phi , 3r^2\sin^2 \theta \sin^2 \phi )$

$(\cot\phi, r^2 \sin\theta\cos^2 \phi , 3r^2\sin^2 \theta \sin^2 \phi )$

$(\tan \phi, r^2 \sin\theta\cos \phi \cos \theta, 3r^2\sin^2 \theta \sin^2 \phi )$

$(\tan \phi, r^2 \sin\theta\cos^2 \theta, 3r^2\sin^2 \theta \sin^2 \phi )$

$(\cot\phi, r^2 \sin\theta\cos^2 \theta, 3r^2\sin^2 \theta \sin^2 \phi )$

Correct answer:

$(\tan \phi, r^2 \sin\theta\cos \phi \cos \theta, 3r^2\sin^2 \theta \sin^2 \phi )$

Explanation:

The conversion from Cartesian to spherical coordinates is as follows:

$x\rightarrow r\sin \theta \cos \phi$

$y\rightarrow r\sin \theta \sin\phi$

$z\rightarrow r\cos \phi$

The three components of the vector therefore become:

$\frac{y}{x}\rightarrow \frac{ r\sin \theta \sin\phi}{ r\sin \theta \cos\phi}=\tan \phi$

$xz\rightarrow( r \sin \theta \cos \phi )(r \cos \theta)=r^2 \sin\theta\cos \phi \cos \theta,$

$3y^2=3r^2 \sin^2 \theta \sin^2 \phi$

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Example Question #2171 : Calculus 3

Find an equation of a sphere that has the line segment joining (0,4,2) and (6,0,2) as a diameter.

Possible Answers:

(x+3)^2+(y-1)^2+(z-2)^2=12

(x-2)^2+(y-3)^2+(z-2)^2=13

(x-3)^2+(y-2)^2+(z-2)^2=13

(x-3)^2+(y-2)^2+(z-2)^2=52

None of the Above.

Correct answer:

(x-3)^2+(y-2)^2+(z-2)^2=13

Explanation:

To find the diameter we need to find the distance between the two points, where:

$D=\sqrt{(6-0)^2+(0-4)^2+(2-2)^2}$

$D=\sqrt{52}=2\sqrt{13}$

Since the radius is half of the Diameter,

$r=\sqrt{13}$

To find the the Center point we use the mid-point formula. So,

$Center = \left ( \frac{0+6}{2},\frac{4+0}{2},\frac{2+2}{2} \right )=\left ( 3,2,2\right )$

So the equation of the sphere is:

(x-3)^2+(y-2)^2+(z-2)^2=13

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Example Question #172 : Spherical Coordinates

Which of the following describes a sphere with radius of 2 and center at the origin together with its interior?

Possible Answers:

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\geq 2\right \}$

None of the Above.

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\geq 4\right \}$

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\leq 4\right \}$

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\leq 2\right \}$

Correct answer:

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\leq 4\right \}$

Explanation:

If the Center of the sphere is the origin and the Radius is 2.

The equations of the sphere is:

(x-0)^2+(y-0)^2+(z-0)^2=2^2

Therefore, the interior of the sphere is defined at:

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\leq 4\right \}$

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Example Question #513 : 3 Dimensional Space

Find a rectangular equation for the surface whose spherical equation is $\rho=\sin\phi \cos\Theta$ .

Possible Answers:

$x^2+(y-\frac{1}{2})^2+z^2=\frac{1}{4}$

$x^2+y^2+z^2=\frac{1}{4}$

$x^2+y^2+(z-\frac{1}{2})^2=\frac{1}{4}$

$(x-\frac{1}{2})^2+y^2+z^2=\frac{1}{4}$

Correct answer:

$(x-\frac{1}{2})^2+y^2+z^2=\frac{1}{4}$

Explanation:

Recall the relationships between rectangular coordinates (x,y,z) and spherical coordinates $(\rho , \Theta , \phi )$ in three-dimensional space:

$x=\rho\ \sin\phi \cos\Theta$ ,

$y=\rho\ \sin\phi \sin\Theta$ ,

$z=\rho \cos\phi$ ,

$\rho^2=x^2+y^2+z^2$ .

We are given the equation $\rho=\sin\phi \cos\Theta$ to convert to rectangular coordinates. From these relationships, we have

$\rho^2=\rho (\sin\phi \cos\Theta)=x^2+y^2+z^2$ .

Substitute for $\rho\ \sin\phi \cos\Theta$ to yield

x=x^2+y^2+z^2 .

Subtract from both sides:

x^2-x+y^2+z^2=0

Complete the square for the variable :

$(x^2-x+\frac{1}{4})+y^2+z^2=0+\frac{1}{4}$

$\Rightarrow (x-\frac{1}{2})^2+y^2+z^2=\frac{1}{4}$

This is the equation for the aforementioned surface in rectangular coordinates. It describes a sphere whose center is located at $(\frac{1}{2},0,0)$ and whose radius is $\frac{1}{2}$ units long.

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Example Question #174 : Spherical Coordinates

Express the three-dimensional (x,y,z) Cartesian coordinates as spherical coordinates (ρ, θ, φ):

$\left ( 2, 1, -2\right )$

Possible Answers:

$\left (9, \tan^{-1}\frac{1}{2},\cos^{-1}\frac{-2}{3} \right)$

$\left (3,\frac{1}{2},\frac{-2}{3} \right)$

$\left (9, \tan^{-1}\frac{1}{2},\pi - \cos^{-1}\frac{2}{3} \right)$

$\left (3, \tan^{-1}\frac{1}{2},\pi - \cos^{-1}\frac{2}{3} \right)$

$\left (3, \tan^{-1}\frac{1}{2}, \cos^{-1}\frac{-2}{3} \right)$

Correct answer:

$\left (3, \tan^{-1}\frac{1}{2},\pi - \cos^{-1}\frac{2}{3} \right)$

Explanation:

The coordinates (2, 1, -2) corresponds to: x = 2, y = 1, z = -2, and are to be converted to the spherical coordinates in form of (ρ, θ, φ), where:

$\\ \rho = \sqrt{x^{2}+y^{2}+z^{2}} \\ \tan \theta = \frac{y}{x} \\ \cos \phi = \frac{z}{\rho}$

So, filling in for x, y, z:

$\\ \rho = \sqrt{2^{2}+1^{2}+\left (-2 \right )^{2}} = \sqrt{9} = 3 \\ \tan \theta = \frac{1}{2} \Rightarrow \theta = \tan^{-1}\frac{1}{2} \\ \cos \phi = \frac{-2}{3} \Rightarrow \phi = \cos^{-1}\frac{-2}{3} \Rightarrow \pi - \cos^{-1}\frac{2}{3}$

Then the spherical coordinates are represented as:

$\left (3, \tan^{-1}\frac{1}{2}, \pi - \cos^{-1}\frac{2}{3} \right )$

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Example Question #175 : Spherical Coordinates

Express the three-dimensional (x,y,z) Cartesian coordinates as spherical coordinates (ρ, θ, φ):

$\left (0, 3, 4 \right)$

Possible Answers:

$\left (5, \frac{3}{0}, \frac{4}{5} \right)$

$\left (25, \frac{1}{2}\pi, \cos^{-1}\frac{4}{5} \right)$

$\left (5, \frac{1}{2}, \cos^{-1}\frac{4}{5} \right)$

$\left (25, \tan^{-1}\frac{1}{2}\pi, \cos^{-1}\frac{4}{5} \right)$

$\left (5, \frac{1}{2}\pi, \cos^{-1}\frac{4}{5} \right)$

Correct answer:

$\left (5, \frac{1}{2}\pi, \cos^{-1}\frac{4}{5} \right)$

Explanation:

The coordinates (0, 3, 4) corresponds to: x = 0, y = 3, z = 4, and are to be converted to the spherical coordinates in form of (ρ, θ, φ), where:

$\\ \rho = \sqrt{x^{2}+y^{2}+z^{2}} \\ \tan \theta = \frac{y}{x} \\ \cos \phi = \frac{z}{\rho}$

So, filling in for x, y, z:

$\\ \rho = \sqrt{0^{2}+3^{2}+ 4^{2}} = \sqrt{9 + 16} = \sqrt{25}= 5 \\ \tan \theta = \frac{3}{0} \Rightarrow \theta = \tan^{-1}\frac{3}{0} \Rightarrow \theta = \frac{1}{2}\pi \\ \cos \phi = \frac{4}{5} \Rightarrow \phi = \cos^{-1}\frac{4}{5}$

Then the spherical coordinates are represented as:

$\left (5, \frac{1}{2}\pi, \cos^{-1}\frac{4}{5} \right)$

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Example Question #176 : Spherical Coordinates

Express the three-dimensional (x,y,z) Cartesian coordinates as spherical coordinates (ρ, θ, φ):

$\left ( \sqrt{2}, 1, 1\right )$

Possible Answers:

$\left (2, \tan^{-1}\frac{\sqrt{2}}{2}, \frac{1}{3}\pi \right)$

$\left (2, \frac{\sqrt{2}}{2}, \frac{1}{3}\pi \right)$

$\left (4, \tan^{-1}\frac{\sqrt{2}}{2}, \cos^{-1}\frac{1}{3}\pi \right)$

$\left (4, \frac{\sqrt{2}}{2}, \frac{1}{3}\pi \right)$

$\left (2, \tan^{-1}\frac{\sqrt{2}}{2}, \cos^{-1}\frac{1}{3}\pi \right)$

Correct answer:

$\left (2, \tan^{-1}\frac{\sqrt{2}}{2}, \frac{1}{3}\pi \right)$

Explanation:

The coordinates (√2, 1, 1) corresponds to: x = √2, y = 1, z = 1, and are to be converted to the spherical coordinates in form of (ρ, θ, φ), where:

$\\ \rho = \sqrt{x^{2}+y^{2}+z^{2}} \\ \tan \theta = \frac{y}{x} \\ \cos \phi = \frac{z}{\rho}$

So, filling in for x, y, z:

$\\ \rho = \sqrt{\sqrt{2}^{2}+1^{2}+ 1^{2}} = \sqrt{2 + 1 + 1} = \sqrt{4}= 2 \\ \tan \theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = \tan^{-1}\frac{1}{\sqrt{2}} \Rightarrow \theta = \tan^{-1}\frac{\sqrt{2}}{2} \\ \cos \phi = \frac{1}{2} \Rightarrow \phi = \cos^{-1}\frac{1}{2} \Rightarrow \phi = \frac{1}{3}\pi$

Then the spherical coordinates are represented as:

$\left (2, \tan^{-1}\frac{\sqrt{2}}{2}, \frac{1}{3}\pi \right)$

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Example Question #177 : Spherical Coordinates

Express the three-dimensional spherical coordinates (ρ, θ, φ) as Cartesian coordinate (x, y, z):

$\left ( 8, \frac{1}{4}\pi, \frac{1}{6}\pi\right )$

Possible Answers:

$\left ( 2\sqrt{2}, 4\sqrt{3}, 2\sqrt{2}\right )$

$\left ( 2\sqrt{2}, \cos^{-1}2\sqrt{2}, \sin^{-1}4\sqrt{3}\right )$

$\left ( 2\sqrt{2}, 2\sqrt{2}, 4\sqrt{3}\right )$

$\left ( \frac{\sqrt{2}}{4}, \frac{\sqrt{2}}{4}, \frac{\sqrt{3}}{2}\right )$

$\left ( 4\sqrt{2}, 4\sqrt{2}, 4\sqrt{3}\right )$

Correct answer:

$\left ( 2\sqrt{2}, 2\sqrt{2}, 4\sqrt{3}\right )$

Explanation:

The coordinates (8, π/4, π/6) corresponds to: ρ = 8, θ = π/4, φ = π/6, and are to be converted to the Cartesian coordinates in form of (x, y, z), where:

$\\ x = \rho\cos\theta\sin\phi \\ y = \rho\sin\theta\sin\phi \\ z = \rho\cos\phi$

So, filling in for ρ, θ, φ:

$\\ x = 8\cos\frac{1}{4}\pi\cdot\sin\frac{1}{6}\pi = 8\cdot\frac{\sqrt{2}}{2}\cdot\frac{1}{2}= 2\sqrt{2} \\ y = 8\sin\frac{1}{4}\pi\cdot\sin\frac{1}{6}\pi = 8\cdot\frac{\sqrt{2}}{2}\cdot\frac{1}{2} = 2\sqrt{2} \\ z = 8\cos\frac{1}{6}\pi = 8\cdot\frac{\sqrt{3}}{2} = 4\sqrt{3}$

Then the Cartesian coordinates are represented as:

$\left ( 2\sqrt{2}, 2\sqrt{2}, 4\sqrt{3}\right )$

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Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #511 : 3 Dimensional Space

Example Question #512 : 3 Dimensional Space

Example Question #513 : 3 Dimensional Space

Example Question #2171 : Calculus 3

Example Question #172 : Spherical Coordinates

Example Question #513 : 3 Dimensional Space

Example Question #174 : Spherical Coordinates

Example Question #175 : Spherical Coordinates

Example Question #176 : Spherical Coordinates

Example Question #177 : Spherical Coordinates

All Calculus 3 Resources

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