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Calculus 3 : 3-Dimensional Space

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Example Questions

Example Question #421 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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 #71 : 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 #72 : Spherical Coordinates

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

Possible Answers:

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

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

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

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

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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Example Question #81 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-134,106,18).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(171.8,141.65^{\circ},83.99^{\circ})$

$(171.8,38.35^{\circ},83.99^{\circ})$

$(106.54,38.35^{\circ},96.01^{\circ})$

$(106.54,141.65^{\circ},83.99^{\circ})$

Correct answer:

$(171.8,141.65^{\circ},83.99^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-134,106,18)\\&\rho=\sqrt{(-134)^2+(106)^2+(18)^2}=171.8\\&\theta=arctan(\frac{106}{-134})=141.65^{\circ}\\&\phi=arccos(\frac{18}{171.8})=83.99^{\circ}\end{align*}$

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Example Question #81 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(129,59,25).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(133.28,24.58^{\circ},80^{\circ})$

$(144.04,155.42^{\circ},80^{\circ})$

$(133.28,155.42^{\circ},100^{\circ})$

$(144.04,24.58^{\circ},80^{\circ})$

Correct answer:

$(144.04,24.58^{\circ},80^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(129,59,25)\\&\rho=\sqrt{(129)^2+(59)^2+(25)^2}=144.04\\&\theta=arctan(\frac{59}{129})=24.58^{\circ}\\&\phi=arccos(\frac{25}{144.04})=80^{\circ}\end{align*}$

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Example Question #81 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(95,114,147).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(176.68,50.19^{\circ},45.27^{\circ})$

$(208.88,129.81^{\circ},45.27^{\circ})$

$(208.88,50.19^{\circ},45.27^{\circ})$

$(176.68,129.81^{\circ},134.73^{\circ})$

Correct answer:

$(208.88,50.19^{\circ},45.27^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(95,114,147)\\&\rho=\sqrt{(95)^2+(114)^2+(147)^2}=208.88\\&\theta=arctan(\frac{114}{95})=50.19^{\circ}\\&\phi=arccos(\frac{147}{208.88})=45.27^{\circ}\end{align*}$

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Example Question #427 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-150,110,34).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(126.09,36.25^{\circ},100.36^{\circ})$

$(126.09,143.75^{\circ},79.64^{\circ})$

$(189.09,36.25^{\circ},79.64^{\circ})$

$(189.09,143.75^{\circ},79.64^{\circ})$

Correct answer:

$(189.09,143.75^{\circ},79.64^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-150,110,34)\\&\rho=\sqrt{(-150)^2+(110)^2+(34)^2}=189.09\\&\theta=arctan(\frac{110}{-150})=143.75^{\circ}\\&\phi=arccos(\frac{34}{189.09})=79.64^{\circ}\end{align*}$

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Example Question #428 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(147,8,-6).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(123,3.12^{\circ},92.33^{\circ})$

$(147.34,176.88^{\circ},92.33^{\circ})$

$(123,176.88^{\circ},87.67^{\circ})$

$(147.34,3.12^{\circ},92.33^{\circ})$

Correct answer:

$(147.34,3.12^{\circ},92.33^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(147,8,-6)\\&\rho=\sqrt{(147)^2+(8)^2+(-6)^2}=147.34\\&\theta=arctan(\frac{8}{147})=3.12^{\circ}\\&\phi=arccos(\frac{-6}{147.34})=92.33^{\circ}\end{align*}$

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Example Question #2092 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(91,-82,-1).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(122.5,-42.02^{\circ},90.47^{\circ})$

$(58.69,137.98^{\circ},89.53^{\circ})$

$(58.69,-42.02^{\circ},90.47^{\circ})$

$(122.5,137.98^{\circ},90.47^{\circ})$

Correct answer:

$(122.5,-42.02^{\circ},90.47^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(91,-82,-1)\\&\rho=\sqrt{(91)^2+(-82)^2+(-1)^2}=122.5\\&\theta=arctan(\frac{-82}{91})=-42.02^{\circ}\\&\phi=arccos(\frac{-1}{122.5})=90.47^{\circ}\end{align*}$

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Example Question #81 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(121,22,104).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(142.72,10.3^{\circ},49.78^{\circ})$

$(161.06,10.3^{\circ},49.78^{\circ})$

$(161.06,169.7^{\circ},49.78^{\circ})$

$(142.72,169.7^{\circ},130.22^{\circ})$

Correct answer:

$(161.06,10.3^{\circ},49.78^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(121,22,104)\\&\rho=\sqrt{(121)^2+(22)^2+(104)^2}=161.06\\&\theta=arctan(\frac{22}{121})=10.3^{\circ}\\&\phi=arccos(\frac{104}{161.06})=49.78^{\circ}\end{align*}$

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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 #421 : 3 Dimensional Space

Example Question #71 : Spherical Coordinates

Example Question #72 : Spherical Coordinates

Example Question #81 : Spherical Coordinates

Example Question #81 : Spherical Coordinates

Example Question #81 : Spherical Coordinates

Example Question #427 : 3 Dimensional Space

Example Question #428 : 3 Dimensional Space

Example Question #2092 : Calculus 3

Example Question #81 : Spherical Coordinates

All Calculus 3 Resources

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