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Calculus 3 : 3-Dimensional Space

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Example Questions

Example Question #411 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-23,-69,-91).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(116.49,-108.43^{\circ},141.37^{\circ})$

$(116.49,71.57^{\circ},141.37^{\circ})$

$(72.73,-108.43^{\circ},141.37^{\circ})$

$(72.73,71.57^{\circ},38.63^{\circ})$

Correct answer:

$(116.49,-108.43^{\circ},141.37^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-23,-69,-91)\\&\rho=\sqrt{(-23)^2+(-69)^2+(-91)^2}=116.49\\&\theta=arctan(\frac{-69}{-23})=-108.43^{\circ}\\&\phi=arccos(\frac{-91}{116.49})=141.37^{\circ}\end{align*}$

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Example Question #412 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(97,-21,117).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(99.25,-12.22^{\circ},40.31^{\circ})$

$(99.25,167.78^{\circ},139.69^{\circ})$

$(153.42,167.78^{\circ},40.31^{\circ})$

$(153.42,-12.22^{\circ},40.31^{\circ})$

Correct answer:

$(153.42,-12.22^{\circ},40.31^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(97,-21,117)\\&\rho=\sqrt{(97)^2+(-21)^2+(117)^2}=153.42\\&\theta=arctan(\frac{-21}{97})=-12.22^{\circ}\\&\phi=arccos(\frac{117}{153.42})=40.31^{\circ}\end{align*}$

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Example Question #61 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-33,81,-31).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(92.8,112.17^{\circ},109.52^{\circ})$

$(87.46,112.17^{\circ},109.52^{\circ})$

$(87.46,67.83^{\circ},70.48^{\circ})$

$(92.8,67.83^{\circ},109.52^{\circ})$

Correct answer:

$(92.8,112.17^{\circ},109.52^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-33,81,-31)\\&\rho=\sqrt{(-33)^2+(81)^2+(-31)^2}=92.8\\&\theta=arctan(\frac{81}{-33})=112.17^{\circ}\\&\phi=arccos(\frac{-31}{92.8})=109.52^{\circ}\end{align*}$

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Example Question #71 : Spherical Coordinates

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

Possible Answers:

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

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

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

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

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 #411 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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 #2081 : 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:

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

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

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

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

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 #411 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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 #412 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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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Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #411 : 3 Dimensional Space

Example Question #412 : 3 Dimensional Space

Example Question #61 : Spherical Coordinates

Example Question #71 : Spherical Coordinates

Example Question #71 : Spherical Coordinates

Example Question #2081 : Calculus 3

Example Question #411 : 3 Dimensional Space

Example Question #2081 : Calculus 3

Example Question #411 : 3 Dimensional Space

Example Question #412 : 3 Dimensional Space

All Calculus 3 Resources

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