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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #172 : Spherical Coordinates

Which of the following describes a sphere with radius of 2 and center at the origin together with its interior?

Possible Answers:

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\geq 2\right \}$

None of the Above.

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\geq 4\right \}$

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\leq 4\right \}$

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\leq 2\right \}$

Correct answer:

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\leq 4\right \}$

Explanation:

If the Center of the sphere is the origin and the Radius is 2.

The equations of the sphere is:

(x-0)^2+(y-0)^2+(z-0)^2=2^2

Therefore, the interior of the sphere is defined at:

$\Omega=\left \{ (x,y,z):x^2+y^2+z^2\leq 4\right \}$

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Example Question #513 : 3 Dimensional Space

Find a rectangular equation for the surface whose spherical equation is $\rho=\sin\phi \cos\Theta$ .

Possible Answers:

$x^2+(y-\frac{1}{2})^2+z^2=\frac{1}{4}$

$x^2+y^2+z^2=\frac{1}{4}$

$x^2+y^2+(z-\frac{1}{2})^2=\frac{1}{4}$

$(x-\frac{1}{2})^2+y^2+z^2=\frac{1}{4}$

Correct answer:

$(x-\frac{1}{2})^2+y^2+z^2=\frac{1}{4}$

Explanation:

Recall the relationships between rectangular coordinates (x,y,z) and spherical coordinates $(\rho , \Theta , \phi )$ in three-dimensional space:

$x=\rho\ \sin\phi \cos\Theta$ ,

$y=\rho\ \sin\phi \sin\Theta$ ,

$z=\rho \cos\phi$ ,

$\rho^2=x^2+y^2+z^2$ .

We are given the equation $\rho=\sin\phi \cos\Theta$ to convert to rectangular coordinates. From these relationships, we have

$\rho^2=\rho (\sin\phi \cos\Theta)=x^2+y^2+z^2$ .

Substitute for $\rho\ \sin\phi \cos\Theta$ to yield

x=x^2+y^2+z^2 .

Subtract from both sides:

x^2-x+y^2+z^2=0

Complete the square for the variable :

$(x^2-x+\frac{1}{4})+y^2+z^2=0+\frac{1}{4}$

$\Rightarrow (x-\frac{1}{2})^2+y^2+z^2=\frac{1}{4}$

This is the equation for the aforementioned surface in rectangular coordinates. It describes a sphere whose center is located at $(\frac{1}{2},0,0)$ and whose radius is $\frac{1}{2}$ units long.

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Example Question #174 : Spherical Coordinates

Express the three-dimensional (x,y,z) Cartesian coordinates as spherical coordinates (ρ, θ, φ):

$\left ( 2, 1, -2\right )$

Possible Answers:

$\left (9, \tan^{-1}\frac{1}{2},\cos^{-1}\frac{-2}{3} \right)$

$\left (3,\frac{1}{2},\frac{-2}{3} \right)$

$\left (9, \tan^{-1}\frac{1}{2},\pi - \cos^{-1}\frac{2}{3} \right)$

$\left (3, \tan^{-1}\frac{1}{2},\pi - \cos^{-1}\frac{2}{3} \right)$

$\left (3, \tan^{-1}\frac{1}{2}, \cos^{-1}\frac{-2}{3} \right)$

Correct answer:

$\left (3, \tan^{-1}\frac{1}{2},\pi - \cos^{-1}\frac{2}{3} \right)$

Explanation:

The coordinates (2, 1, -2) corresponds to: x = 2, y = 1, z = -2, and are to be converted to the spherical coordinates in form of (ρ, θ, φ), where:

$\\ \rho = \sqrt{x^{2}+y^{2}+z^{2}} \\ \tan \theta = \frac{y}{x} \\ \cos \phi = \frac{z}{\rho}$

So, filling in for x, y, z:

$\\ \rho = \sqrt{2^{2}+1^{2}+\left (-2 \right )^{2}} = \sqrt{9} = 3 \\ \tan \theta = \frac{1}{2} \Rightarrow \theta = \tan^{-1}\frac{1}{2} \\ \cos \phi = \frac{-2}{3} \Rightarrow \phi = \cos^{-1}\frac{-2}{3} \Rightarrow \pi - \cos^{-1}\frac{2}{3}$

Then the spherical coordinates are represented as:

$\left (3, \tan^{-1}\frac{1}{2}, \pi - \cos^{-1}\frac{2}{3} \right )$

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Example Question #175 : Spherical Coordinates

Express the three-dimensional (x,y,z) Cartesian coordinates as spherical coordinates (ρ, θ, φ):

$\left (0, 3, 4 \right)$

Possible Answers:

$\left (5, \frac{3}{0}, \frac{4}{5} \right)$

$\left (25, \frac{1}{2}\pi, \cos^{-1}\frac{4}{5} \right)$

$\left (5, \frac{1}{2}, \cos^{-1}\frac{4}{5} \right)$

$\left (25, \tan^{-1}\frac{1}{2}\pi, \cos^{-1}\frac{4}{5} \right)$

$\left (5, \frac{1}{2}\pi, \cos^{-1}\frac{4}{5} \right)$

Correct answer:

$\left (5, \frac{1}{2}\pi, \cos^{-1}\frac{4}{5} \right)$

Explanation:

The coordinates (0, 3, 4) corresponds to: x = 0, y = 3, z = 4, and are to be converted to the spherical coordinates in form of (ρ, θ, φ), where:

$\\ \rho = \sqrt{x^{2}+y^{2}+z^{2}} \\ \tan \theta = \frac{y}{x} \\ \cos \phi = \frac{z}{\rho}$

So, filling in for x, y, z:

$\\ \rho = \sqrt{0^{2}+3^{2}+ 4^{2}} = \sqrt{9 + 16} = \sqrt{25}= 5 \\ \tan \theta = \frac{3}{0} \Rightarrow \theta = \tan^{-1}\frac{3}{0} \Rightarrow \theta = \frac{1}{2}\pi \\ \cos \phi = \frac{4}{5} \Rightarrow \phi = \cos^{-1}\frac{4}{5}$

Then the spherical coordinates are represented as:

$\left (5, \frac{1}{2}\pi, \cos^{-1}\frac{4}{5} \right)$

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Example Question #176 : Spherical Coordinates

Express the three-dimensional (x,y,z) Cartesian coordinates as spherical coordinates (ρ, θ, φ):

$\left ( \sqrt{2}, 1, 1\right )$

Possible Answers:

$\left (2, \tan^{-1}\frac{\sqrt{2}}{2}, \frac{1}{3}\pi \right)$

$\left (2, \frac{\sqrt{2}}{2}, \frac{1}{3}\pi \right)$

$\left (4, \tan^{-1}\frac{\sqrt{2}}{2}, \cos^{-1}\frac{1}{3}\pi \right)$

$\left (4, \frac{\sqrt{2}}{2}, \frac{1}{3}\pi \right)$

$\left (2, \tan^{-1}\frac{\sqrt{2}}{2}, \cos^{-1}\frac{1}{3}\pi \right)$

Correct answer:

$\left (2, \tan^{-1}\frac{\sqrt{2}}{2}, \frac{1}{3}\pi \right)$

Explanation:

The coordinates (√2, 1, 1) corresponds to: x = √2, y = 1, z = 1, and are to be converted to the spherical coordinates in form of (ρ, θ, φ), where:

$\\ \rho = \sqrt{x^{2}+y^{2}+z^{2}} \\ \tan \theta = \frac{y}{x} \\ \cos \phi = \frac{z}{\rho}$

So, filling in for x, y, z:

$\\ \rho = \sqrt{\sqrt{2}^{2}+1^{2}+ 1^{2}} = \sqrt{2 + 1 + 1} = \sqrt{4}= 2 \\ \tan \theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = \tan^{-1}\frac{1}{\sqrt{2}} \Rightarrow \theta = \tan^{-1}\frac{\sqrt{2}}{2} \\ \cos \phi = \frac{1}{2} \Rightarrow \phi = \cos^{-1}\frac{1}{2} \Rightarrow \phi = \frac{1}{3}\pi$

Then the spherical coordinates are represented as:

$\left (2, \tan^{-1}\frac{\sqrt{2}}{2}, \frac{1}{3}\pi \right)$

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Example Question #177 : Spherical Coordinates

Express the three-dimensional spherical coordinates (ρ, θ, φ) as Cartesian coordinate (x, y, z):

$\left ( 8, \frac{1}{4}\pi, \frac{1}{6}\pi\right )$

Possible Answers:

$\left ( 2\sqrt{2}, 4\sqrt{3}, 2\sqrt{2}\right )$

$\left ( 2\sqrt{2}, \cos^{-1}2\sqrt{2}, \sin^{-1}4\sqrt{3}\right )$

$\left ( 2\sqrt{2}, 2\sqrt{2}, 4\sqrt{3}\right )$

$\left ( \frac{\sqrt{2}}{4}, \frac{\sqrt{2}}{4}, \frac{\sqrt{3}}{2}\right )$

$\left ( 4\sqrt{2}, 4\sqrt{2}, 4\sqrt{3}\right )$

Correct answer:

$\left ( 2\sqrt{2}, 2\sqrt{2}, 4\sqrt{3}\right )$

Explanation:

The coordinates (8, π/4, π/6) corresponds to: ρ = 8, θ = π/4, φ = π/6, and are to be converted to the Cartesian coordinates in form of (x, y, z), where:

$\\ x = \rho\cos\theta\sin\phi \\ y = \rho\sin\theta\sin\phi \\ z = \rho\cos\phi$

So, filling in for ρ, θ, φ:

$\\ x = 8\cos\frac{1}{4}\pi\cdot\sin\frac{1}{6}\pi = 8\cdot\frac{\sqrt{2}}{2}\cdot\frac{1}{2}= 2\sqrt{2} \\ y = 8\sin\frac{1}{4}\pi\cdot\sin\frac{1}{6}\pi = 8\cdot\frac{\sqrt{2}}{2}\cdot\frac{1}{2} = 2\sqrt{2} \\ z = 8\cos\frac{1}{6}\pi = 8\cdot\frac{\sqrt{3}}{2} = 4\sqrt{3}$

Then the Cartesian coordinates are represented as:

$\left ( 2\sqrt{2}, 2\sqrt{2}, 4\sqrt{3}\right )$

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Example Question #178 : Spherical Coordinates

Express the three-dimensional (x,y,z) Cartesian coordinates as spherical coordinates (ρ, θ, φ):

$\left ( 2\sqrt{3},6,-4\right )$

Possible Answers:

$\left ( 8,\tan^{-1}\frac{1}{3}\pi, \cos^{-1}\frac{2}{3}\pi\right )$

$\left ( 64,\tan^{-1}\frac{1}{3}\pi, \cos^{-1}\frac{2}{3}\pi\right )$

$\left ( 64,\frac{1}{3}\pi, \frac{2}{3}\pi\right )$

$\left ( 8,\frac{2}{3}\pi,\frac{1}{3}\pi\right )$

$\left ( 8,\frac{1}{3}\pi, \frac{2}{3}\pi\right )$

Correct answer:

$\left ( 8,\frac{1}{3}\pi, \frac{2}{3}\pi\right )$

Explanation:

The coordinates (2√3, 6, -4) corresponds to: x = 2√3, y = 6, z = -4, and are to be converted to the spherical coordinates in form of (ρ, θ, φ), where:

$\\ \rho = \sqrt{x^{2}+y^{2}+z^{2}} \\ \tan \theta = \frac{y}{x} \\ \cos \phi = \frac{z}{\rho}$

So, filling in for x, y, z:

$\\ \rho = \sqrt{\left ( 2\sqrt{3}\right )^{2}+6^{2}+ \left ( -4\right )^{2}} = \sqrt{4\cdot3 + 36 + 16} = \sqrt{64}= 8 \\ \tan \theta = \frac{6}{2\sqrt{3}} \Rightarrow \theta = \tan^{-1}\frac{6}{2\sqrt{3}} \Rightarrow \theta = \tan^{-1}\frac{\sqrt{3}}{1} \Rightarrow \theta = \frac{1}{3}\pi \\ \cos \phi = \frac{-4}{8} \Rightarrow \phi = \cos^{-1}\frac{-1}{2} \Rightarrow \phi = \pi - \frac{1}{3}\pi = \frac{2}{3}\pi$

Then the spherical coordinates are represented as:

$\left ( 8,\frac{1}{3}\pi, \frac{2}{3}\pi\right )$

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Example Question #2181 : Calculus 3

Calculate the following Integral.

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx$

Possible Answers:

$\frac{1}{120}$

120

$-\frac{3}{120}$

$-\frac{1}{120}$

Correct answer:

$-\frac{3}{120}$

Explanation:

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx=\int_{0}^{1}\Big(\int_{x}^{x^2} xy^2 dy\Big)dx$

Lets deal with the inner integral first.

$\int_{x}^{x^2}xy^2 dy$

$\int_{x}^{x^2}xy^2 dy=\frac{1}{3}y^3x\Big|_{x}^{x^2}$

$=(\frac{1}{3}(x^2)^3\cdot x)-(\frac{1}{3}x^3\cdot x)$

$=\frac{1}{3}x^7-\frac{1}{3}x^4$

Now we evaluate this expression in the outer integral.

$\int_{0}^{1} \frac{1}{3}x^7-\frac{1}{3}x^4dx$

$=\frac{1}{3}\int_{0}^{1} x^7-x^4dx$

$=\frac{1}{3}\Big( \frac{1}{8}x^8-\frac{1}{5}x^5\Big)\Big|_{0}^{1}$

$=\frac{1}{3}\Big(\frac{1}{8}1^8-\frac{1}{5}1^5\Big)-\frac{1}{3}\Big(\frac{1}{8}0^8-\frac{1}{5}0^5\Big)$

$=\frac{1}{3}\Big(\frac{1}{8}-\frac{1}{5}\Big)-\frac{1}{3}\Big(0-0\Big)$

$=\frac{1}{3}\Big(\frac{5}{40}-\frac{8}{40}\Big)$

$=\frac{1}{3}\Big(-\frac{3}{40}\Big)=-\frac{3}{120}$

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Example Question #1 : Double Integration Over General Regions

Calculate the definite integral of the function f(x,y) , given below as

$f(x,y)= 3x^3\sqrt{y}+ \sin(2x)\cos(4y)$

Possible Answers:

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{8}\cos(2x)\sin(4y)$

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}+\frac{1}{8}\cos(2x)\sin(4y)+C$

Cannot be solved.

$\iint f(x,y)dxdy=x^4y^{3/2}-\cos(2x)\sin(4y)+C$

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{8}\cos(2x)\sin(4y)+C$

Correct answer:

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{8}\cos(2x)\sin(4y)+C$

Explanation:

Because there are no nested terms containing both and , we can rewrite the integral as

$\iint f(x,y)dxdy=\int3x^3dx \int \sqrt{y}dy+\int\sin(2x)dx \int \cos(4y)dy$

This enables us to evaluate the double integral and the product of two independent single integrals. From the integration rules from single-variable calculus, we should arrive at the result

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{12}\cos(2x)\sin(4y)+C$ .

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Example Question #2182 : Calculus 3

Evaluate the following integral on the region specified:

$\int\int_R(x^3y)\:dxdy$

Where R is the region defined by the conditions:

$0\leq x \leq 2, \; 1\leq y \leq2$

Possible Answers:

7.5

$4\sqrt{2}$

$2\sqrt{2}$

Correct answer:

Explanation:

$\int^2_1\int^2_0(x^3y)\:dxdy=\int^2_1(\frac{x^4y}{4}|^2_0)dy\int^2_14y\:dy =2y^2|^2_1=2(4-1)=2(3)=6$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #172 : Spherical Coordinates

Example Question #513 : 3 Dimensional Space

Example Question #174 : Spherical Coordinates

Example Question #175 : Spherical Coordinates

Example Question #176 : Spherical Coordinates

Example Question #177 : Spherical Coordinates

Example Question #178 : Spherical Coordinates

Example Question #2181 : Calculus 3

Example Question #1 : Double Integration Over General Regions

Example Question #2182 : Calculus 3

All Calculus 3 Resources

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