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Calculus 3 : 3-Dimensional Space

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

Example Questions

Example Question #41 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (17,11,37) . What is the position of this point in spherical coordinates?

Possible Answers:

$(20.25,147.09^{\circ},-151.31^{\circ})$

$(20.25,32.91^{\circ},28.69^{\circ})$

$(20.25,147.09^{\circ},151.31^{\circ})$

$(42.18,32.91^{\circ},28.69^{\circ})$

$(42.18,147.09^{\circ},28.69^{\circ})$

Correct answer:

$(42.18,32.91^{\circ},28.69^{\circ})$

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

$\begin{align*}&(17,11,37)\\&\rho=\sqrt{(17)^2+(11)^2+(37)^2}=42.18\\&\theta=arctan(\frac{11}{17})=32.91^{\circ}\\&\phi=arccos(\frac{37}{42.18})=28.69^{\circ}\end{align*}$

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Example Question #42 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-18,-14,35) . What is the position of this point in spherical coordinates?

Possible Answers:

$(22.8,37.87^{\circ},146.91^{\circ})$

$(41.77,37.87^{\circ},33.09^{\circ})$

$(22.8,37.87^{\circ},-146.91^{\circ})$

$(41.77,-142.13^{\circ},33.09^{\circ})$

$(22.8,-142.13^{\circ},33.09^{\circ})$

Correct answer:

$(41.77,-142.13^{\circ},33.09^{\circ})$

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

$\begin{align*}&(-18,-14,35)\\&\rho=\sqrt{(-18)^2+(-14)^2+(35)^2}=41.77\\&\theta=arctan(\frac{-14}{-18})=-142.13^{\circ}\\&\phi=arccos(\frac{35}{41.77})=33.09^{\circ}\end{align*}$

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Example Question #41 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(25,12,111).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(27.73,154.36^{\circ},165.97^{\circ})$

$(27.73,154.36^{\circ},-165.97^{\circ})$

$(114.41,154.36^{\circ},14.03^{\circ})$

$(27.73,25.64^{\circ},14.03^{\circ})$

$(114.41,25.64^{\circ},14.03^{\circ})$

Correct answer:

$(114.41,25.64^{\circ},14.03^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent to be careful when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(25,12,111)\\&\rho=\sqrt{(25)^2+(12)^2+(111)^2}=114.41\\&\theta=arctan(\frac{12}{25})=25.64^{\circ}\\&\phi=arccos(\frac{111}{114.41})=14.03^{\circ}\end{align*}$

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Example Question #51 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-71,-55,-115).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(89.81,37.76^{\circ},37.99^{\circ})$

$(145.91,-142.24^{\circ},142.01^{\circ})$

$(89.81,-142.24^{\circ},142.01^{\circ})$

$(145.91,37.76^{\circ},142.01^{\circ})$

$(89.81,37.76^{\circ},-37.99^{\circ})$

Correct answer:

$(145.91,-142.24^{\circ},142.01^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-71,-55,-115)\\&\rho=\sqrt{(-71)^2+(-55)^2+(-115)^2}=145.91\\&\theta=arctan(\frac{-55}{-71})=-142.24^{\circ}\\&\phi=arccos(\frac{-115}{145.91})=142.01^{\circ}\end{align*}$

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Example Question #52 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(132,44,-6).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(139.14,161.57^{\circ},87.53^{\circ})$

$(139.14,161.57^{\circ},-87.53^{\circ})$

$(139.27,18.43^{\circ},92.47^{\circ})$

$(139.27,161.57^{\circ},92.47^{\circ})$

$(139.14,18.43^{\circ},92.47^{\circ})$

Correct answer:

$(139.27,18.43^{\circ},92.47^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(132,44,-6)\\&\rho=\sqrt{(132)^2+(44)^2+(-6)^2}=139.27\\&\theta=arctan(\frac{44}{132})=18.43^{\circ}\\&\phi=arccos(\frac{-6}{139.27})=92.47^{\circ}\end{align*}$

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Example Question #51 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(13,67,7).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(68.25,100.98^{\circ},95.86^{\circ})$

$(68.25,100.98^{\circ},-95.86^{\circ})$

$(68.61,100.98^{\circ},84.14^{\circ})$

$(68.61,79.02^{\circ},84.14^{\circ})$

$(68.25,79.02^{\circ},84.14^{\circ})$

Correct answer:

$(68.61,79.02^{\circ},84.14^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(13,67,7)\\&\rho=\sqrt{(13)^2+(67)^2+(7)^2}=68.61\\&\theta=arctan(\frac{67}{13})=79.02^{\circ}\\&\phi=arccos(\frac{7}{68.61})=84.14^{\circ}\end{align*}$

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Example Question #52 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(149,-85,-119).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(208.77,150.3^{\circ},124.75^{\circ})$

$(171.54,-29.7^{\circ},124.75^{\circ})$

$(208.77,-29.7^{\circ},124.75^{\circ})$

$(171.54,150.3^{\circ},55.25^{\circ})$

$(171.54,150.3^{\circ},-55.25^{\circ})$

Correct answer:

$(208.77,-29.7^{\circ},124.75^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(149,-85,-119)\\&\rho=\sqrt{(149)^2+(-85)^2+(-119)^2}=208.77\\&\theta=arctan(\frac{-85}{149})=-29.7^{\circ}\\&\phi=arccos(\frac{-119}{208.77})=124.75^{\circ}\end{align*}$

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Example Question #55 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-117,-131,-29).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(178.02,-131.77^{\circ},99.38^{\circ})$

$(175.64,-131.77^{\circ},99.38^{\circ})$

$(178.02,48.23^{\circ},99.38^{\circ})$

$(175.64,48.23^{\circ},80.62^{\circ})$

$(175.64,48.23^{\circ},-80.62^{\circ})$

Correct answer:

$(178.02,-131.77^{\circ},99.38^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-117,-131,-29)\\&\rho=\sqrt{(-117)^2+(-131)^2+(-29)^2}=178.02\\&\theta=arctan(\frac{-131}{-117})=-131.77^{\circ}\\&\phi=arccos(\frac{-29}{178.02})=99.38^{\circ}\end{align*}$

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Example Question #56 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-16,-40,79).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(43.08,68.2^{\circ},-151.39^{\circ})$

$(89.98,-111.8^{\circ},28.61^{\circ})$

$(43.08,-111.8^{\circ},28.61^{\circ})$

$(89.98,68.2^{\circ},28.61^{\circ})$

$(43.08,68.2^{\circ},151.39^{\circ})$

Correct answer:

$(89.98,-111.8^{\circ},28.61^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-16,-40,79)\\&\rho=\sqrt{(-16)^2+(-40)^2+(79)^2}=89.98\\&\theta=arctan(\frac{-40}{-16})=-111.8^{\circ}\\&\phi=arccos(\frac{79}{89.98})=28.61^{\circ}\end{align*}$

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Example Question #51 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(38,82,130).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(158.33,65.14^{\circ},34.81^{\circ})$

$(90.38,114.86^{\circ},-145.19^{\circ})$

$(90.38,114.86^{\circ},145.19^{\circ})$

$(90.38,65.14^{\circ},34.81^{\circ})$

$(158.33,114.86^{\circ},34.81^{\circ})$

Correct answer:

$(158.33,65.14^{\circ},34.81^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(38,82,130)\\&\rho=\sqrt{(38)^2+(82)^2+(130)^2}=158.33\\&\theta=arctan(\frac{82}{38})=65.14^{\circ}\\&\phi=arccos(\frac{130}{158.33})=34.81^{\circ}\end{align*}$

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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 #41 : Spherical Coordinates

Example Question #42 : Spherical Coordinates

Example Question #41 : Spherical Coordinates

Example Question #51 : Spherical Coordinates

Example Question #52 : Spherical Coordinates

Example Question #51 : Spherical Coordinates

Example Question #52 : Spherical Coordinates

Example Question #55 : Spherical Coordinates

Example Question #56 : Spherical Coordinates

Example Question #51 : Spherical Coordinates

All Calculus 3 Resources

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