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Calculus 3 : 3-Dimensional Space

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Example Questions

Example Question #431 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(72,26,-76).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(36.38,160.14^{\circ},45.21^{\circ})$

$(36.38,19.86^{\circ},134.79^{\circ})$

$(107.87,160.14^{\circ},134.79^{\circ})$

$(107.87,19.86^{\circ},134.79^{\circ})$

Correct answer:

$(107.87,19.86^{\circ},134.79^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(72,26,-76)\\&\rho=\sqrt{(72)^2+(26)^2+(-76)^2}=107.87\\&\theta=arctan(\frac{26}{72})=19.86^{\circ}\\&\phi=arccos(\frac{-76}{107.87})=134.79^{\circ}\end{align*}$

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Example Question #2097 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(50,-125,38).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(121.04,-68.2^{\circ},74.24^{\circ})$

$(121.04,111.8^{\circ},105.76^{\circ})$

$(139.89,111.8^{\circ},74.24^{\circ})$

$(139.89,-68.2^{\circ},74.24^{\circ})$

Correct answer:

$(139.89,-68.2^{\circ},74.24^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(50,-125,38)\\&\rho=\sqrt{(50)^2+(-125)^2+(38)^2}=139.89\\&\theta=arctan(\frac{-125}{50})=-68.2^{\circ}\\&\phi=arccos(\frac{38}{139.89})=74.24^{\circ}\end{align*}$

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Example Question #2098 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(48,69,118).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(144.88,55.18^{\circ},35.46^{\circ})$

$(144.88,124.82^{\circ},35.46^{\circ})$

$(127.69,124.82^{\circ},144.54^{\circ})$

$(127.69,55.18^{\circ},35.46^{\circ})$

Correct answer:

$(144.88,55.18^{\circ},35.46^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(48,69,118)\\&\rho=\sqrt{(48)^2+(69)^2+(118)^2}=144.88\\&\theta=arctan(\frac{69}{48})=55.18^{\circ}\\&\phi=arccos(\frac{118}{144.88})=35.46^{\circ}\end{align*}$

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Example Question #2099 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(145,81,25).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(167.96,29.19^{\circ},81.44^{\circ})$

$(153.2,29.19^{\circ},81.44^{\circ})$

$(153.2,150.81^{\circ},98.56^{\circ})$

$(167.96,150.81^{\circ},81.44^{\circ})$

Correct answer:

$(167.96,29.19^{\circ},81.44^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(145,81,25)\\&\rho=\sqrt{(145)^2+(81)^2+(25)^2}=167.96\\&\theta=arctan(\frac{81}{145})=29.19^{\circ}\\&\phi=arccos(\frac{25}{167.96})=81.44^{\circ}\end{align*}$

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Example Question #431 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(129,24,-145).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(195.56,169.46^{\circ},137.86^{\circ})$

$(96.12,169.46^{\circ},42.14^{\circ})$

$(195.56,10.54^{\circ},137.86^{\circ})$

$(96.12,10.54^{\circ},137.86^{\circ})$

Correct answer:

$(195.56,10.54^{\circ},137.86^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(129,24,-145)\\&\rho=\sqrt{(129)^2+(24)^2+(-145)^2}=195.56\\&\theta=arctan(\frac{24}{129})=10.54^{\circ}\\&\phi=arccos(\frac{-145}{195.56})=137.86^{\circ}\end{align*}$

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Example Question #432 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-114,109,-5).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(157.8,136.28^{\circ},91.82^{\circ})$

$(57.15,136.28^{\circ},91.82^{\circ})$

$(57.15,43.72^{\circ},88.18^{\circ})$

$(157.8,43.72^{\circ},91.82^{\circ})$

Correct answer:

$(157.8,136.28^{\circ},91.82^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-114,109,-5)\\&\rho=\sqrt{(-114)^2+(109)^2+(-5)^2}=157.8\\&\theta=arctan(\frac{109}{-114})=136.28^{\circ}\\&\phi=arccos(\frac{-5}{157.8})=91.82^{\circ}\end{align*}$

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Example Question #433 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(104,-87,16).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(77.77,140.09^{\circ},96.73^{\circ})$

$(136.53,-39.91^{\circ},83.27^{\circ})$

$(77.77,-39.91^{\circ},83.27^{\circ})$

$(136.53,140.09^{\circ},83.27^{\circ})$

Correct answer:

$(136.53,-39.91^{\circ},83.27^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(104,-87,16)\\&\rho=\sqrt{(104)^2+(-87)^2+(16)^2}=136.53\\&\theta=arctan(\frac{-87}{104})=-39.91^{\circ}\\&\phi=arccos(\frac{16}{136.53})=83.27^{\circ}\end{align*}$

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Example Question #434 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(39,-141,35).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(150.42,105.46^{\circ},76.55^{\circ})$

$(150.42,-74.54^{\circ},76.55^{\circ})$

$(139.27,-74.54^{\circ},76.55^{\circ})$

$(139.27,105.46^{\circ},103.45^{\circ})$

Correct answer:

$(150.42,-74.54^{\circ},76.55^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(39,-141,35)\\&\rho=\sqrt{(39)^2+(-141)^2+(35)^2}=150.42\\&\theta=arctan(\frac{-141}{39})=-74.54^{\circ}\\&\phi=arccos(\frac{35}{150.42})=76.55^{\circ}\end{align*}$

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Example Question #435 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-41,-136,-3).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(142.08,73.22^{\circ},91.21^{\circ})$

$(142.08,-106.78^{\circ},91.21^{\circ})$

$(137.23,-106.78^{\circ},91.21^{\circ})$

$(137.23,73.22^{\circ},88.79^{\circ})$

Correct answer:

$(142.08,-106.78^{\circ},91.21^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-41,-136,-3)\\&\rho=\sqrt{(-41)^2+(-136)^2+(-3)^2}=142.08\\&\theta=arctan(\frac{-136}{-41})=-106.78^{\circ}\\&\phi=arccos(\frac{-3}{142.08})=91.21^{\circ}\end{align*}$

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Example Question #436 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-93,-113,-89).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(171.29,50.55^{\circ},121.31^{\circ})$

$(143.46,-129.45^{\circ},121.31^{\circ})$

$(143.46,50.55^{\circ},58.69^{\circ})$

$(171.29,-129.45^{\circ},121.31^{\circ})$

Correct answer:

$(171.29,-129.45^{\circ},121.31^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-93,-113,-89)\\&\rho=\sqrt{(-93)^2+(-113)^2+(-89)^2}=171.29\\&\theta=arctan(\frac{-113}{-93})=-129.45^{\circ}\\&\phi=arccos(\frac{-89}{171.29})=121.31^{\circ}\end{align*}$

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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 #431 : 3 Dimensional Space

Example Question #2097 : Calculus 3

Example Question #2098 : Calculus 3

Example Question #2099 : Calculus 3

Example Question #431 : 3 Dimensional Space

Example Question #432 : 3 Dimensional Space

Example Question #433 : 3 Dimensional Space

Example Question #434 : 3 Dimensional Space

Example Question #435 : 3 Dimensional Space

Example Question #436 : 3 Dimensional Space

All Calculus 3 Resources

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