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Calculus 3 : Spherical Coordinates

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Example Questions

Example Question #71 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(93,77,-37).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(126.28,140.38^{\circ},107.04^{\circ})$

$(120.74,39.62^{\circ},107.04^{\circ})$

$(120.74,140.38^{\circ},72.96^{\circ})$

$(126.28,39.62^{\circ},107.04^{\circ})$

Correct answer:

$(126.28,39.62^{\circ},107.04^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(93,77,-37)\\&\rho=\sqrt{(93)^2+(77)^2+(-37)^2}=126.28\\&\theta=arctan(\frac{77}{93})=39.62^{\circ}\\&\phi=arccos(\frac{-37}{126.28})=107.04^{\circ}\end{align*}$

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Example Question #71 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-85,87,135).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(181.71,134.33^{\circ},42.02^{\circ})$

$(121.63,134.33^{\circ},42.02^{\circ})$

$(121.63,45.67^{\circ},137.98^{\circ})$

$(181.71,45.67^{\circ},42.02^{\circ})$

Correct answer:

$(181.71,134.33^{\circ},42.02^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-85,87,135)\\&\rho=\sqrt{(-85)^2+(87)^2+(135)^2}=181.71\\&\theta=arctan(\frac{87}{-85})=134.33^{\circ}\\&\phi=arccos(\frac{135}{181.71})=42.02^{\circ}\end{align*}$

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Example Question #2081 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-52,52,-18).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(73.54,45^{\circ},76.25^{\circ})$

$(75.71,135^{\circ},103.75^{\circ})$

$(75.71,45^{\circ},103.75^{\circ})$

$(73.54,135^{\circ},103.75^{\circ})$

Correct answer:

$(75.71,135^{\circ},103.75^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-52,52,-18)\\&\rho=\sqrt{(-52)^2+(52)^2+(-18)^2}=75.71\\&\theta=arctan(\frac{52}{-52})=135^{\circ}\\&\phi=arccos(\frac{-18}{75.71})=103.75^{\circ}\end{align*}$

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Example Question #2081 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(100,81,-100).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(81,39.01^{\circ},127.85^{\circ})$

$(162.98,140.99^{\circ},127.85^{\circ})$

$(81,140.99^{\circ},52.15^{\circ})$

$(162.98,39.01^{\circ},127.85^{\circ})$

Correct answer:

$(162.98,39.01^{\circ},127.85^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(100,81,-100)\\&\rho=\sqrt{(100)^2+(81)^2+(-100)^2}=162.98\\&\theta=arctan(\frac{81}{100})=39.01^{\circ}\\&\phi=arccos(\frac{-100}{162.98})=127.85^{\circ}\end{align*}$

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Example Question #2084 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(116,26,-104).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(76.84,12.63^{\circ},131.18^{\circ})$

$(157.95,167.37^{\circ},131.18^{\circ})$

$(157.95,12.63^{\circ},131.18^{\circ})$

$(76.84,167.37^{\circ},48.82^{\circ})$

Correct answer:

$(157.95,12.63^{\circ},131.18^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(116,26,-104)\\&\rho=\sqrt{(116)^2+(26)^2+(-104)^2}=157.95\\&\theta=arctan(\frac{26}{116})=12.63^{\circ}\\&\phi=arccos(\frac{-104}{157.95})=131.18^{\circ}\end{align*}$

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Example Question #2085 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-90,-28,75).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(120.45,17.28^{\circ},51.49^{\circ})$

$(120.45,-162.72^{\circ},51.49^{\circ})$

$(69.04,-162.72^{\circ},51.49^{\circ})$

$(69.04,17.28^{\circ},128.51^{\circ})$

Correct answer:

$(120.45,-162.72^{\circ},51.49^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-90,-28,75)\\&\rho=\sqrt{(-90)^2+(-28)^2+(75)^2}=120.45\\&\theta=arctan(\frac{-28}{-90})=-162.72^{\circ}\\&\phi=arccos(\frac{75}{120.45})=51.49^{\circ}\end{align*}$

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Example Question #72 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(98,87,-55).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(142.12,41.6^{\circ},112.77^{\circ})$

$(112.75,41.6^{\circ},112.77^{\circ})$

$(142.12,138.4^{\circ},112.77^{\circ})$

$(112.75,138.4^{\circ},67.23^{\circ})$

Correct answer:

$(142.12,41.6^{\circ},112.77^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(98,87,-55)\\&\rho=\sqrt{(98)^2+(87)^2+(-55)^2}=142.12\\&\theta=arctan(\frac{87}{98})=41.6^{\circ}\\&\phi=arccos(\frac{-55}{142.12})=112.77^{\circ}\end{align*}$

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Example Question #71 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(10,-123,-117).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(170.05,-85.35^{\circ},133.47^{\circ})$

$(151.27,-85.35^{\circ},133.47^{\circ})$

$(151.27,94.65^{\circ},46.53^{\circ})$

$(170.05,94.65^{\circ},133.47^{\circ})$

Correct answer:

$(170.05,-85.35^{\circ},133.47^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(10,-123,-117)\\&\rho=\sqrt{(10)^2+(-123)^2+(-117)^2}=170.05\\&\theta=arctan(\frac{-123}{10})=-85.35^{\circ}\\&\phi=arccos(\frac{-117}{170.05})=133.47^{\circ}\end{align*}$

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Example Question #72 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-109,54,-1).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(121.65,153.65^{\circ},90.47^{\circ})$

$(121.65,26.35^{\circ},90.47^{\circ})$

$(104.39,153.65^{\circ},90.47^{\circ})$

$(104.39,26.35^{\circ},89.53^{\circ})$

Correct answer:

$(121.65,153.65^{\circ},90.47^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-109,54,-1)\\&\rho=\sqrt{(-109)^2+(54)^2+(-1)^2}=121.65\\&\theta=arctan(\frac{54}{-109})=153.65^{\circ}\\&\phi=arccos(\frac{-1}{121.65})=90.47^{\circ}\end{align*}$

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Example Question #2087 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-93,-2,-106).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(141.03,1.23^{\circ},138.73^{\circ})$

$(141.03,-178.77^{\circ},138.73^{\circ})$

$(125.9,-178.77^{\circ},138.73^{\circ})$

$(125.9,1.23^{\circ},41.27^{\circ})$

Correct answer:

$(141.03,-178.77^{\circ},138.73^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-93,-2,-106)\\&\rho=\sqrt{(-93)^2+(-2)^2+(-106)^2}=141.03\\&\theta=arctan(\frac{-2}{-93})=-178.77^{\circ}\\&\phi=arccos(\frac{-106}{141.03})=138.73^{\circ}\end{align*}$

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Calculus 3 : Spherical Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #71 : Spherical Coordinates

Example Question #71 : Spherical Coordinates

Example Question #2081 : Calculus 3

Example Question #2081 : Calculus 3

Example Question #2084 : Calculus 3

Example Question #2085 : Calculus 3

Example Question #72 : Spherical Coordinates

Example Question #71 : Spherical Coordinates

Example Question #72 : Spherical Coordinates

Example Question #2087 : Calculus 3

All Calculus 3 Resources

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