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Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #251 : Cylindrical Coordinates

Express the three-dimensional Cartesian coordinates (x,y,z) as three-dimensional cylindrical coordinates (r, θ, z):

$\left ( -2, 2, 3\right )$

Possible Answers:

$\left ( 8,-1, 3\right )$

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

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

$\left ( 8, \frac{3}{4}\pi, \sqrt{3}\right )$

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

Correct answer:

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

Explanation:

The coordinates (-2, 2, 3) corresponds to: x = -2, y = 2, z = 3, and are to be converted to the cylindrical coordinates in form of (r, θ, z), where:

$\\ r = \sqrt{x^{2}+y^{2}} \\ \tan(\theta) = \frac{y}{x} \\ z = z$

So, filling in for x, y, z:

$\\ r = {\sqrt{\left ( -2\right )^{2}+2^{2}}} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2} \\ \tan(\theta) = \frac{2}{-2} \Rightarrow \theta = \tan^{-1}\ \left ( -1\right ) \Rightarrow \theta = \frac{3}{4}\pi \\ z = 3$

Then the cylindrical coordinates are represented as:

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

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Example Question #252 : Cylindrical Coordinates

Express the three-dimensional cylindrical coordinates (r, θ, z) as three-dimensional (x,y,z) Cartesian coordinates:

$\left ( 1, 45^{\circ}, 1\right )$

Possible Answers:

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

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

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

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

$\left ( \frac{45\sqrt{2}}{2}, 1,\frac{45\sqrt{2}}{2}\right )$

Correct answer:

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

Explanation:

The coordinates (1, 45°, 1) corresponds to: r = 1, θ = 45°, z = 1, and are to be converted to the Cartesian coordinates in form of (x, y, z), where:

$\\ x = r\cos\theta \\ y = r\sin\theta \\ z = z$

So, filling in for r, θ, z:

$\\ x = 1\cos{45^{\circ}} = 1\cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\ y = 1\sin{45^{\circ}} = 1\cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\ z = 1$

Then the Cartesian coordinates are represented as:

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

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Example Question #253 : Cylindrical Coordinates

Express the three-dimensional cylindrical coordinates (r, θ, z) as three-dimensional (x,y,z) Cartesian coordinates:

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

Possible Answers:

$\left ( 2,0,-4\right )$

$\left (\cos^{-1}0,\sin^{-1}2,-4\right )$

$\left ( 2,0,16\right )$

$\left (\cos^{-1}0,\sin^{-1}2,16\right )$

$\left ( 0,2,-4\right )$

Correct answer:

$\left ( 0,2,-4\right )$

Explanation:

The coordinates (2, π/2, -4) corresponds to: r = 2, θ = π/2, z = -4, and are to be converted to the Cartesian coordinates in form of (x, y, z), where:

$\\ x = r\cos\theta \\ y = r\sin\theta \\ z = z$

So, filling in for r, θ, z:

$\\ x = 2\cos{\frac{1}{2}\pi} = 2\cdot 0 = 0 \\ y = 2\sin{\frac{1}{2}\pi} = 2\cdot 1 = 2 \\ z = -4$

Then the Cartesian coordinates are represented as:

$\left ( 0,2,-4\right )$

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Example Question #1 : Spherical Coordinates

Convert the following into spherical coordinates.

$x=\sin(t)$

$y=\cos(t)$

z=1

Possible Answers:

$\rho=\sqrt{2}$

$\phi=\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$

$\rho=\sqrt{2}$

$\phi=0$

$\theta=\tan(\cot(t))$

$\rho=2$

$\phi=-\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$

$\rho=1$

$\phi=\frac{\pi}{2}$

$\theta=\tan(\cot(t))$

$\rho=\sqrt{2}$

$\phi=\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$

Correct answer:

$\rho=\sqrt{2}$

$\phi=\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$

Explanation:

In order to convert to spherical coordinates , we need to remember the conversion equations.

$\rho=\sqrt{x^2+y^2+z^2}$

$\theta=\tan^{-1}\Big(\frac{y}{x}\Big)$

$\phi=\tan^{-1}\Big(\frac{\sqrt{x^2+y^2}}{z}\Big)$

Now lets apply this to our problem.

$\rho=\sqrt{(\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{1+1}=\sqrt{2}$

$\phi=\tan^{-1}\Big(\frac{\sqrt{(\sin{t})^2+(\cos{t})^2}}{1}\Big)=\tan^{-1}(1)=\frac{\pi}{4}$

$\theta=\tan^{-1}\Big(\frac{\cos(t)}{\sin(t)}\Big)=\tan^{-1}(\cot(t))$

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Example Question #2 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (9,3,2) . What is the position of this point in spherical coordinates?

Possible Answers:

$(9.49,161.57^{\circ},78.1^{\circ})$

$(9.7,18.43^{\circ},78.1^{\circ})$

$(9.49,18.43^{\circ},78.1^{\circ})$

$(9.49,161.57^{\circ},-101.9^{\circ})$

$(9.7,161.57^{\circ},78.1^{\circ})$

Correct answer:

$(9.7,18.43^{\circ},78.1^{\circ})$

Explanation:

When given Cartesian coordinates of the form (x,y,z) to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(9,3,2)\\&\rho=\sqrt{(9)^2+(3)^2+(2)^2}=9.7\\&\theta=arctan(\frac{3}{9})=18.43^{\circ}\\&\phi=arccos(\frac{2}{9.7})=78.1^{\circ}\end{align*}$

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Example Question #1 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-12,4,-3) . What is the position of this point in spherical coordinates?

Possible Answers:

$(12.65,18.43^{\circ},76.66^{\circ})$

$(12.65,18.43^{\circ},-76.66^{\circ})$

$(13,161.57^{\circ},103.34^{\circ})$

$(13,18.43^{\circ},103.34^{\circ})$

$(12.65,161.57^{\circ},103.34^{\circ})$

Correct answer:

$(13,161.57^{\circ},103.34^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(-12,4,-3)\\&\rho=\sqrt{(-12)^2+(4)^2+(-3)^2}=13\\&\theta=arctan(\frac{4}{-12})=161.57^{\circ}\\&\phi=arccos(\frac{-3}{13})=103.34^{\circ}\end{align*}$

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Example Question #3 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (5,1,-6) . What is the position of this point in spherical coordinates?

Possible Answers:

$(7.87,168.69^{\circ},139.64^{\circ})$

$(5.1,168.69^{\circ},-40.36^{\circ})$

$(7.87,11.31^{\circ},139.64^{\circ})$

$(5.1,168.69^{\circ},40.36^{\circ})$

$(5.1,11.31^{\circ},139.64^{\circ})$

Correct answer:

$(7.87,11.31^{\circ},139.64^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(5,1,-6)\\&\rho=\sqrt{(5)^2+(1)^2+(-6)^2}=7.87\\&\theta=arctan(\frac{1}{5})=11.31^{\circ}\\&\phi=arccos(\frac{-6}{7.87})=139.64^{\circ}\end{align*}$

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Example Question #341 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (5,-8,3) . What is the position of this point in spherical coordinates?

Possible Answers:

$(9.43,122.01^{\circ},-107.64^{\circ})$

$(9.43,122.01^{\circ},107.64^{\circ})$

$(9.9,122.01^{\circ},72.36^{\circ})$

$(9.9,-57.99^{\circ},72.36^{\circ})$

$(9.43,-57.99^{\circ},72.36^{\circ})$

Correct answer:

$(9.9,-57.99^{\circ},72.36^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(5,-8,3)\\&\rho=\sqrt{(5)^2+(-8)^2+(3)^2}=9.9\\&\theta=arctan(\frac{-8}{5})=-57.99^{\circ}\\&\phi=arccos(\frac{3}{9.9})=72.36^{\circ}\end{align*}$

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Example Question #6 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-9,-9,-9) . What is the position of this point in spherical coordinates?

Possible Answers:

$(12.73,45^{\circ},54.74^{\circ})$

$(15.59,-135^{\circ},125.26^{\circ})$

$(15.59,45^{\circ},125.26^{\circ})$

$(12.73,45^{\circ},-54.74^{\circ})$

$(12.73,-135^{\circ},125.26^{\circ})$

Correct answer:

$(15.59,-135^{\circ},125.26^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(-9,-9,-9)\\&\rho=\sqrt{(-9)^2+(-9)^2+(-9)^2}=15.59\\&\theta=arctan(\frac{-9}{-9})=-135^{\circ}\\&\phi=arccos(\frac{-9}{15.59})=125.26^{\circ}\end{align*}$

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Example Question #5 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (11,0,-5) . What is the position of this point in spherical coordinates?

Possible Answers:

$(11,180^{\circ},65.56^{\circ})$

$(12.08,180^{\circ},114.44^{\circ})$

$(11,180^{\circ},-65.56^{\circ})$

$(11,0^{\circ},114.44^{\circ})$

$(12.08,0^{\circ},114.44^{\circ})$

Correct answer:

$(12.08,0^{\circ},114.44^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(11,0,-5)\\&\rho=\sqrt{(11)^2+(0)^2+(-5)^2}=12.08\\&\theta=arctan(\frac{0}{11})=0^{\circ}\\&\phi=arccos(\frac{-5}{12.08})=114.44^{\circ}\end{align*}$

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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 #251 : Cylindrical Coordinates

Example Question #252 : Cylindrical Coordinates

Example Question #253 : Cylindrical Coordinates

Example Question #1 : Spherical Coordinates

Example Question #2 : Spherical Coordinates

Example Question #1 : Spherical Coordinates

Example Question #3 : Spherical Coordinates

Example Question #341 : 3 Dimensional Space

Example Question #6 : Spherical Coordinates

Example Question #5 : Spherical Coordinates

All Calculus 3 Resources

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