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Calculus 3 : 3-Dimensional Space

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Example Questions

Example Question #441 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(35,-8,-45).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(57.57,167.12^{\circ},141.42^{\circ})$

$(36.53,-12.88^{\circ},141.42^{\circ})$

$(36.53,167.12^{\circ},38.58^{\circ})$

$(57.57,-12.88^{\circ},141.42^{\circ})$

Correct answer:

$(57.57,-12.88^{\circ},141.42^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(35,-8,-45)\\&\rho=\sqrt{(35)^2+(-8)^2+(-45)^2}=57.57\\&\theta=arctan(\frac{-8}{35})=-12.88^{\circ}\\&\phi=arccos(\frac{-45}{57.57})=141.42^{\circ}\end{align*}$

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Example Question #441 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(100,26,15).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(100.69,165.43^{\circ},98.26^{\circ})$

$(104.41,165.43^{\circ},81.74^{\circ})$

$(104.41,14.57^{\circ},81.74^{\circ})$

$(100.69,14.57^{\circ},81.74^{\circ})$

Correct answer:

$(104.41,14.57^{\circ},81.74^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(100,26,15)\\&\rho=\sqrt{(100)^2+(26)^2+(15)^2}=104.41\\&\theta=arctan(\frac{26}{100})=14.57^{\circ}\\&\phi=arccos(\frac{15}{104.41})=81.74^{\circ}\end{align*}$

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Example Question #442 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(126,-64,77).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(160.94,153.07^{\circ},61.42^{\circ})$

$(160.94,-26.93^{\circ},61.42^{\circ})$

$(129.96,153.07^{\circ},118.58^{\circ})$

$(129.96,-26.93^{\circ},61.42^{\circ})$

Correct answer:

$(160.94,-26.93^{\circ},61.42^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(126,-64,77)\\&\rho=\sqrt{(126)^2+(-64)^2+(77)^2}=160.94\\&\theta=arctan(\frac{-64}{126})=-26.93^{\circ}\\&\phi=arccos(\frac{77}{160.94})=61.42^{\circ}\end{align*}$

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Example Question #443 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(76,-36,20).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(86.44,154.65^{\circ},76.62^{\circ})$

$(73.7,154.65^{\circ},103.38^{\circ})$

$(86.44,-25.35^{\circ},76.62^{\circ})$

$(73.7,-25.35^{\circ},76.62^{\circ})$

Correct answer:

$(86.44,-25.35^{\circ},76.62^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(76,-36,20)\\&\rho=\sqrt{(76)^2+(-36)^2+(20)^2}=86.44\\&\theta=arctan(\frac{-36}{76})=-25.35^{\circ}\\&\phi=arccos(\frac{20}{86.44})=76.62^{\circ}\end{align*}$

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Example Question #2111 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-128,-134,9).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(165.13,46.31^{\circ},92.78^{\circ})$

$(165.13,-133.69^{\circ},87.22^{\circ})$

$(185.53,-133.69^{\circ},87.22^{\circ})$

$(185.53,46.31^{\circ},87.22^{\circ})$

Correct answer:

$(185.53,-133.69^{\circ},87.22^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-128,-134,9)\\&\rho=\sqrt{(-128)^2+(-134)^2+(9)^2}=185.53\\&\theta=arctan(\frac{-134}{-128})=-133.69^{\circ}\\&\phi=arccos(\frac{9}{185.53})=87.22^{\circ}\end{align*}$

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Example Question #2112 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(84,131,-111).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(191.15,57.33^{\circ},125.5^{\circ})$

$(113.78,122.67^{\circ},54.5^{\circ})$

$(113.78,57.33^{\circ},125.5^{\circ})$

$(191.15,122.67^{\circ},125.5^{\circ})$

Correct answer:

$(191.15,57.33^{\circ},125.5^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(84,131,-111)\\&\rho=\sqrt{(84)^2+(131)^2+(-111)^2}=191.15\\&\theta=arctan(\frac{131}{84})=57.33^{\circ}\\&\phi=arccos(\frac{-111}{191.15})=125.5^{\circ}\end{align*}$

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Example Question #2113 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(21,-9,-147).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(146.87,156.8^{\circ},-8.83^{\circ})$

$(148.76,156.8^{\circ},171.17^{\circ})$

$(146.87,156.8^{\circ},8.83^{\circ})$

$(148.76,-23.2^{\circ},171.17^{\circ})$

Correct answer:

$(148.76,-23.2^{\circ},171.17^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(21,-9,-147)\\&\rho=\sqrt{(21)^2+(-9)^2+(-147)^2}=148.76\\&\theta=arctan(\frac{-9}{21})=-23.2^{\circ}\\&\phi=arccos(\frac{-147}{148.76})=171.17^{\circ}\end{align*}$

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Example Question #442 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-49,-102,89).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(143.97,64.34^{\circ},51.81^{\circ})$

$(77.96,64.34^{\circ},128.19^{\circ})$

$(143.97,-115.66^{\circ},51.81^{\circ})$

$(77.96,64.34^{\circ},-128.19^{\circ})$

Correct answer:

$(143.97,-115.66^{\circ},51.81^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-49,-102,89)\\&\rho=\sqrt{(-49)^2+(-102)^2+(89)^2}=143.97\\&\theta=arctan(\frac{-102}{-49})=-115.66^{\circ}\\&\phi=arccos(\frac{89}{143.97})=51.81^{\circ}\end{align*}$

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Example Question #2115 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-57,9,-101).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(106.7,8.97^{\circ},-29.74^{\circ})$

$(116.32,8.97^{\circ},150.26^{\circ})$

$(116.32,171.03^{\circ},150.26^{\circ})$

$(106.7,8.97^{\circ},29.74^{\circ})$

Correct answer:

$(116.32,171.03^{\circ},150.26^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-57,9,-101)\\&\rho=\sqrt{(-57)^2+(9)^2+(-101)^2}=116.32\\&\theta=arctan(\frac{9}{-57})=171.03^{\circ}\\&\phi=arccos(\frac{-101}{116.32})=150.26^{\circ}\end{align*}$

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Example Question #2116 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(57,75,-15).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(84.52,127.23^{\circ},-80.95^{\circ})$

$(95.39,127.23^{\circ},99.05^{\circ})$

$(84.52,127.23^{\circ},80.95^{\circ})$

$(95.39,52.77^{\circ},99.05^{\circ})$

Correct answer:

$(95.39,52.77^{\circ},99.05^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(57,75,-15)\\&\rho=\sqrt{(57)^2+(75)^2+(-15)^2}=95.39\\&\theta=arctan(\frac{75}{57})=52.77^{\circ}\\&\phi=arccos(\frac{-15}{95.39})=99.05^{\circ}\end{align*}$

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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 #441 : 3 Dimensional Space

Example Question #441 : 3 Dimensional Space

Example Question #442 : 3 Dimensional Space

Example Question #443 : 3 Dimensional Space

Example Question #2111 : Calculus 3

Example Question #2112 : Calculus 3

Example Question #2113 : Calculus 3

Example Question #442 : 3 Dimensional Space

Example Question #2115 : Calculus 3

Example Question #2116 : Calculus 3

All Calculus 3 Resources

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