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Calculus 3 : Spherical Coordinates

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Example Questions

Example Question #394 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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

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 #395 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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

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 #396 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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

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 #397 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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

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 #398 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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

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 #399 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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

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 #400 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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

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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Example Question #401 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(142,-93,-109).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(169.74,146.78^{\circ},-57.29^{\circ})$

$(201.73,-33.22^{\circ},122.71^{\circ})$

$(201.73,146.78^{\circ},122.71^{\circ})$

$(169.74,146.78^{\circ},57.29^{\circ})$

$(169.74,-33.22^{\circ},122.71^{\circ})$

Correct answer:

$(201.73,-33.22^{\circ},122.71^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(142,-93,-109)\\&\rho=\sqrt{(142)^2+(-93)^2+(-109)^2}=201.73\\&\theta=arctan(\frac{-93}{142})=-33.22^{\circ}\\&\phi=arccos(\frac{-109}{201.73})=122.71^{\circ}\end{align*}$

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Example Question #402 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(59,-122,8).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(135.75,115.81^{\circ},86.62^{\circ})$

$(135.52,115.81^{\circ},93.38^{\circ})$

$(135.52,-64.19^{\circ},86.62^{\circ})$

$(135.75,-64.19^{\circ},86.62^{\circ})$

$(135.52,115.81^{\circ},-93.38^{\circ})$

Correct answer:

$(135.75,-64.19^{\circ},86.62^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(59,-122,8)\\&\rho=\sqrt{(59)^2+(-122)^2+(8)^2}=135.75\\&\theta=arctan(\frac{-122}{59})=-64.19^{\circ}\\&\phi=arccos(\frac{8}{135.75})=86.62^{\circ}\end{align*}$

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Example Question #403 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(9,109,-5).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(109.49,94.72^{\circ},92.62^{\circ})$

$(109.49,85.28^{\circ},92.62^{\circ})$

$(109.37,85.28^{\circ},92.62^{\circ})$

$(109.37,94.72^{\circ},-87.38^{\circ})$

$(109.37,94.72^{\circ},87.38^{\circ})$

Correct answer:

$(109.49,85.28^{\circ},92.62^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(9,109,-5)\\&\rho=\sqrt{(9)^2+(109)^2+(-5)^2}=109.49\\&\theta=arctan(\frac{109}{9})=85.28^{\circ}\\&\phi=arccos(\frac{-5}{109.49})=92.62^{\circ}\end{align*}$

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Calculus 3 : Spherical Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #394 : 3 Dimensional Space

Example Question #395 : 3 Dimensional Space

Example Question #396 : 3 Dimensional Space

Example Question #397 : 3 Dimensional Space

Example Question #398 : 3 Dimensional Space

Example Question #399 : 3 Dimensional Space

Example Question #400 : 3 Dimensional Space

Example Question #401 : 3 Dimensional Space

Example Question #402 : 3 Dimensional Space

Example Question #403 : 3 Dimensional Space

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