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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 #11 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-44,-37,-7) . What is the position of this point in spherical coordinates?

Possible Answers:

$(57.49,40.06^{\circ},-83.06^{\circ})$

$(57.91,40.06^{\circ},96.94^{\circ})$

$(57.91,-139.94^{\circ},96.94^{\circ})$

$(57.49,40.06^{\circ},83.06^{\circ})$

$(57.49,-139.94^{\circ},96.94^{\circ})$

Correct answer:

$(57.91,-139.94^{\circ},96.94^{\circ})$

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

$\begin{align*}&(-44,-37,-7)\\&\rho=\sqrt{(-44)^2+(-37)^2+(-7)^2}=57.91\\&\theta=arctan(\frac{-37}{-44})=-139.94^{\circ}\\&\phi=arccos(\frac{-7}{57.91})=96.94^{\circ}\end{align*}$

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Example Question #12 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-24,-34,24) . What is the position of this point in spherical coordinates?

Possible Answers:

$(41.62,54.78^{\circ},-119.97^{\circ})$

$(48.04,-125.22^{\circ},60.03^{\circ})$

$(48.04,54.78^{\circ},60.03^{\circ})$

$(41.62,-125.22^{\circ},60.03^{\circ})$

$(41.62,54.78^{\circ},119.97^{\circ})$

Correct answer:

$(48.04,-125.22^{\circ},60.03^{\circ})$

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

$\begin{align*}&(-24,-34,24)\\&\rho=\sqrt{(-24)^2+(-34)^2+(24)^2}=48.04\\&\theta=arctan(\frac{-34}{-24})=-125.22^{\circ}\\&\phi=arccos(\frac{24}{48.04})=60.03^{\circ}\end{align*}$

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Example Question #13 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (34,43,-21) . What is the position of this point in spherical coordinates?

Possible Answers:

$(58.7,51.67^{\circ},110.96^{\circ})$

$(54.82,128.33^{\circ},69.04^{\circ})$

$(58.7,128.33^{\circ},110.96^{\circ})$

$(54.82,51.67^{\circ},110.96^{\circ})$

$(54.82,128.33^{\circ},-69.04^{\circ})$

Correct answer:

$(58.7,51.67^{\circ},110.96^{\circ})$

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

$\begin{align*}&(34,43,-21)\\&\rho=\sqrt{(34)^2+(43)^2+(-21)^2}=58.7\\&\theta=arctan(\frac{43}{34})=51.67^{\circ}\\&\phi=arccos(\frac{-21}{58.7})=110.96^{\circ}\end{align*}$

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Example Question #21 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-19,-21,-13) . What is the position of this point in spherical coordinates?

Possible Answers:

$(28.32,47.86^{\circ},65.34^{\circ})$

$(28.32,-132.14^{\circ},114.66^{\circ})$

$(31.16,-132.14^{\circ},114.66^{\circ})$

$(28.32,47.86^{\circ},-65.34^{\circ})$

$(31.16,47.86^{\circ},114.66^{\circ})$

Correct answer:

$(31.16,-132.14^{\circ},114.66^{\circ})$

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

$\begin{align*}&(-19,-21,-13)\\&\rho=\sqrt{(-19)^2+(-21)^2+(-13)^2}=31.16\\&\theta=arctan(\frac{-21}{-19})=-132.14^{\circ}\\&\phi=arccos(\frac{-13}{31.16})=114.66^{\circ}\end{align*}$

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Example Question #22 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (0,19,2) . What is the position of this point in spherical coordinates?

Possible Answers:

$(19,-90^{\circ},96.01^{\circ})$

$(19.1,-90^{\circ},83.99^{\circ})$

$(19,90^{\circ},-96.01^{\circ})$

$(19.1,90^{\circ},83.99^{\circ})$

$(19,90^{\circ},83.99^{\circ})$

Correct answer:

$(19.1,90^{\circ},83.99^{\circ})$

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

$\begin{align*}&(0,19,2)\\&\rho=\sqrt{(0)^2+(19)^2+(2)^2}=19.1\\&\theta=arctan(\frac{19}{0})=90^{\circ}\\&\phi=arccos(\frac{2}{19.1})=83.99^{\circ}\end{align*}$

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Example Question #23 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-15,-32,-44) . What is the position of this point in spherical coordinates?

Possible Answers:

$(56.44,64.89^{\circ},141.23^{\circ})$

$(35.34,-115.11^{\circ},141.23^{\circ})$

$(35.34,64.89^{\circ},38.77^{\circ})$

$(35.34,64.89^{\circ},-38.77^{\circ})$

$(56.44,-115.11^{\circ},141.23^{\circ})$

Correct answer:

$(56.44,-115.11^{\circ},141.23^{\circ})$

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

$\begin{align*}&(-15,-32,-44)\\&\rho=\sqrt{(-15)^2+(-32)^2+(-44)^2}=56.44\\&\theta=arctan(\frac{-32}{-15})=-115.11^{\circ}\\&\phi=arccos(\frac{-44}{56.44})=141.23^{\circ}\end{align*}$

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Example Question #361 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (12,-26,-14) . What is the position of this point in spherical coordinates?

Possible Answers:

$(28.64,114.78^{\circ},63.95^{\circ})$

$(28.64,114.78^{\circ},-63.95^{\circ})$

$(31.87,114.78^{\circ},116.05^{\circ})$

$(31.87,-65.22^{\circ},116.05^{\circ})$

$(28.64,-65.22^{\circ},116.05^{\circ})$

Correct answer:

$(31.87,-65.22^{\circ},116.05^{\circ})$

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

$\begin{align*}&(12,-26,-14)\\&\rho=\sqrt{(12)^2+(-26)^2+(-14)^2}=31.87\\&\theta=arctan(\frac{-26}{12})=-65.22^{\circ}\\&\phi=arccos(\frac{-14}{31.87})=116.05^{\circ}\end{align*}$

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Example Question #25 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (36,-33,-14) . What is the position of this point in spherical coordinates?

Possible Answers:

$(48.84,137.49^{\circ},74^{\circ})$

$(48.84,137.49^{\circ},-74^{\circ})$

$(48.84,-42.51^{\circ},106^{\circ})$

$(50.8,137.49^{\circ},106^{\circ})$

$(50.8,-42.51^{\circ},106^{\circ})$

Correct answer:

$(50.8,-42.51^{\circ},106^{\circ})$

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

$\begin{align*}&(36,-33,-14)\\&\rho=\sqrt{(36)^2+(-33)^2+(-14)^2}=50.8\\&\theta=arctan(\frac{-33}{36})=-42.51^{\circ}\\&\phi=arccos(\frac{-14}{50.8})=106^{\circ}\end{align*}$

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Example Question #26 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-30,-33,26) . What is the position of this point in spherical coordinates?

Possible Answers:

$(44.6,-132.27^{\circ},59.76^{\circ})$

$(51.62,-132.27^{\circ},59.76^{\circ})$

$(44.6,47.73^{\circ},120.24^{\circ})$

$(51.62,47.73^{\circ},59.76^{\circ})$

$(44.6,47.73^{\circ},-120.24^{\circ})$

Correct answer:

$(51.62,-132.27^{\circ},59.76^{\circ})$

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

$\begin{align*}&(-30,-33,26)\\&\rho=\sqrt{(-30)^2+(-33)^2+(26)^2}=51.62\\&\theta=arctan(\frac{-33}{-30})=-132.27^{\circ}\\&\phi=arccos(\frac{26}{51.62})=59.76^{\circ}\end{align*}$

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Example Question #27 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-37,39,-35) . What is the position of this point in spherical coordinates?

Possible Answers:

$(64.15,-46.51^{\circ},123.07^{\circ})$

$(53.76,46.51^{\circ},-56.93^{\circ})$

$(64.15,133.49^{\circ},123.07^{\circ})$

$(53.76,-46.51^{\circ},56.93^{\circ})$

$(53.76,133.49^{\circ},123.07^{\circ})$

Correct answer:

$(64.15,133.49^{\circ},123.07^{\circ})$

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

Report an Error

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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 #11 : Spherical Coordinates

Example Question #12 : Spherical Coordinates

Example Question #13 : Spherical Coordinates

Example Question #21 : Spherical Coordinates

Example Question #22 : Spherical Coordinates

Example Question #23 : Spherical Coordinates

Example Question #361 : 3 Dimensional Space

Example Question #25 : Spherical Coordinates

Example Question #26 : Spherical Coordinates

Example Question #27 : Spherical Coordinates

All Calculus 3 Resources

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