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Calculus 3 : Spherical Coordinates

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Example Questions

Example Question #111 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(140,-103,142).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(165.27,-36.34^{\circ},50.75^{\circ})$

$(224.44,143.66^{\circ},129.25^{\circ})$

$(224.44,-36.34^{\circ},50.75^{\circ})$

$(165.27,143.66^{\circ},129.25^{\circ})$

Correct answer:

$(224.44,-36.34^{\circ},50.75^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(140,-103,142)\\&\rho=\sqrt{(140)^2+(-103)^2+(142)^2}=224.44\\&\theta=arctan(\frac{-103}{140})=-36.34^{\circ}\\&\phi=arccos(\frac{142}{224.44})=50.75^{\circ}\end{align*}$

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Example Question #455 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(138,-4,90).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(164.8,178.34^{\circ},123.1^{\circ})$

$(164.8,-1.66^{\circ},56.9^{\circ})$

$(149.73,-1.66^{\circ},56.9^{\circ})$

$(149.73,178.34^{\circ},123.1^{\circ})$

Correct answer:

$(164.8,-1.66^{\circ},56.9^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(138,-4,90)\\&\rho=\sqrt{(138)^2+(-4)^2+(90)^2}=164.8\\&\theta=arctan(\frac{-4}{138})=-1.66^{\circ}\\&\phi=arccos(\frac{90}{164.8})=56.9^{\circ}\end{align*}$

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Example Question #456 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(88,138,47).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(150.57,57.48^{\circ},73.98^{\circ})$

$(170.29,57.48^{\circ},73.98^{\circ})$

$(150.57,122.52^{\circ},106.02^{\circ})$

$(170.29,122.52^{\circ},106.02^{\circ})$

Correct answer:

$(170.29,57.48^{\circ},73.98^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(88,138,47)\\&\rho=\sqrt{(88)^2+(138)^2+(47)^2}=170.29\\&\theta=arctan(\frac{138}{88})=57.48^{\circ}\\&\phi=arccos(\frac{47}{170.29})=73.98^{\circ}\end{align*}$

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Example Question #457 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-140,105,131).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(218.6,143.13^{\circ},53.18^{\circ})$

$(218.6,36.87^{\circ},126.82^{\circ})$

$(87.14,36.87^{\circ},126.82^{\circ})$

$(87.14,143.13^{\circ},53.18^{\circ})$

Correct answer:

$(218.6,143.13^{\circ},53.18^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-140,105,131)\\&\rho=\sqrt{(-140)^2+(105)^2+(131)^2}=218.6\\&\theta=arctan(\frac{105}{-140})=143.13^{\circ}\\&\phi=arccos(\frac{131}{218.6})=53.18^{\circ}\end{align*}$

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Example Question #458 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(54,78,73).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(100.7,55.3^{\circ},52.42^{\circ})$

$(100.7,124.7^{\circ},127.58^{\circ})$

$(119.7,55.3^{\circ},52.42^{\circ})$

$(119.7,124.7^{\circ},127.58^{\circ})$

Correct answer:

$(119.7,55.3^{\circ},52.42^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(54,78,73)\\&\rho=\sqrt{(54)^2+(78)^2+(73)^2}=119.7\\&\theta=arctan(\frac{78}{54})=55.3^{\circ}\\&\phi=arccos(\frac{73}{119.7})=52.42^{\circ}\end{align*}$

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Example Question #459 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-32,47,-99).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(96.52,55.75^{\circ},29.87^{\circ})$

$(114.17,124.25^{\circ},150.13^{\circ})$

$(96.52,124.25^{\circ},150.13^{\circ})$

$(114.17,55.75^{\circ},29.87^{\circ})$

Correct answer:

$(114.17,124.25^{\circ},150.13^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-32,47,-99)\\&\rho=\sqrt{(-32)^2+(47)^2+(-99)^2}=114.17\\&\theta=arctan(\frac{47}{-32})=124.25^{\circ}\\&\phi=arccos(\frac{-99}{114.17})=150.13^{\circ}\end{align*}$

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Example Question #460 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(62,-141,-67).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(142.04,113.74^{\circ},66.49^{\circ})$

$(167.97,113.74^{\circ},66.49^{\circ})$

$(167.97,-66.26^{\circ},113.51^{\circ})$

$(142.04,-66.26^{\circ},113.51^{\circ})$

Correct answer:

$(167.97,-66.26^{\circ},113.51^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(62,-141,-67)\\&\rho=\sqrt{(62)^2+(-141)^2+(-67)^2}=167.97\\&\theta=arctan(\frac{-141}{62})=-66.26^{\circ}\\&\phi=arccos(\frac{-67}{167.97})=113.51^{\circ}\end{align*}$

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Example Question #461 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-137,-121,97).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(206.93,41.45^{\circ},117.95^{\circ})$

$(150.81,-138.55^{\circ},62.05^{\circ})$

$(150.81,41.45^{\circ},117.95^{\circ})$

$(206.93,-138.55^{\circ},62.05^{\circ})$

Correct answer:

$(206.93,-138.55^{\circ},62.05^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-137,-121,97)\\&\rho=\sqrt{(-137)^2+(-121)^2+(97)^2}=206.93\\&\theta=arctan(\frac{-121}{-137})=-138.55^{\circ}\\&\phi=arccos(\frac{97}{206.93})=62.05^{\circ}\end{align*}$

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Example Question #462 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(59,-55,136).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(158.12,137.01^{\circ},149.33^{\circ})$

$(136.7,137.01^{\circ},149.33^{\circ})$

$(136.7,-42.99^{\circ},30.67^{\circ})$

$(158.12,-42.99^{\circ},30.67^{\circ})$

Correct answer:

$(158.12,-42.99^{\circ},30.67^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(59,-55,136)\\&\rho=\sqrt{(59)^2+(-55)^2+(136)^2}=158.12\\&\theta=arctan(\frac{-55}{59})=-42.99^{\circ}\\&\phi=arccos(\frac{136}{158.12})=30.67^{\circ}\end{align*}$

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Example Question #463 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-140,-18,-36).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(145.67,-172.67^{\circ},104.31^{\circ})$

$(145.67,7.33^{\circ},75.69^{\circ})$

$(140.89,-172.67^{\circ},104.31^{\circ})$

$(140.89,7.33^{\circ},75.69^{\circ})$

Correct answer:

$(145.67,-172.67^{\circ},104.31^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-140,-18,-36)\\&\rho=\sqrt{(-140)^2+(-18)^2+(-36)^2}=145.67\\&\theta=arctan(\frac{-18}{-140})=-172.67^{\circ}\\&\phi=arccos(\frac{-36}{145.67})=104.31^{\circ}\end{align*}$

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Calculus 3 : Spherical Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #111 : Spherical Coordinates

Example Question #455 : 3 Dimensional Space

Example Question #456 : 3 Dimensional Space

Example Question #457 : 3 Dimensional Space

Example Question #458 : 3 Dimensional Space

Example Question #459 : 3 Dimensional Space

Example Question #460 : 3 Dimensional Space

Example Question #461 : 3 Dimensional Space

Example Question #462 : 3 Dimensional Space

Example Question #463 : 3 Dimensional Space

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