17Calculus - Triple Integrals in Spherical Coordinates

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On this page we cover triple integrals in spherical coordinates and several applications.
Before going through the material on this page, make sure you understand spherical coordinates and how to convert between spherical and rectangular coordinates. See the spherical coordinates page for detailed explanation and practice problems.

Setting Up and Evaluating Triple Integrals in Spherical Coordinates

Okay, so now you know how to convert an equation or a point to spherical coordinates. So how do we set up a triple integral in spherical coordinates? If you are thinking ahead you probably are anticipating that the $$dV$$ term has at least one extra term like we had in cylindrical coordinates and you would be right. The $$dV$$ term in spherical coordinates has two extra terms, $$\rho^2~\sin\phi$$. So $$dV=\rho^2~\sin\phi~d\rho~d\phi~d\theta$$. Remember that the order of $$d\rho~d\phi~d\theta$$ depends on the order of integration and there are six possible orders. This is just one of them.

In integral form, triple integrals in spherical coordinates look like this.

$$\displaystyle{ \iiint\limits_V {f(x,y,z) ~ dV} = }$$ $$\displaystyle{ \iiint\limits_V{ f(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi ) ~ \rho^2~\sin\phi~d\rho~d\phi~d\theta } }$$

VERY IMPORTANT NOTE - - - Do not forget $$\rho^2~\sin\phi$$ in $$\color{red}{\rho^2~\sin\phi}~d\rho~d\phi~d\theta$$ in the above equation. This is the most common mistake made by students learning this technique.

The next step is to describe the volume in spherical coordinates or, in terms of the integral above, determining the limits of integration. The same techniques apply that you used when setting up triple integrals in rectangular coordinates.

Applications of Triple Integrals in Spherical Coordinates

Just as with rectangular and cylindrical coordinates, the meaning of the function f will determine what is being calculated with the triple integral. Here is a review of a couple of examples.

f

$$\iiint_V{f~dV}$$

1

volume

volume density

mass

Also, physical properties like moment of inertia, center of gravity and force can be calculated using triple integrals.

Okay, time for some practice problems.

Practice

Unless otherwise instructed, evaluate these integrals in the spherical coordinate system.

Basic

Evaluate $$\displaystyle{ \iiint\limits_V { x^2 + y^2 + z^2 ~dV } }$$ where V is the unit ball $$x^2 + y^2 + z^2 \leq 1$$, using spherical coordinates.

Problem Statement

Evaluate $$\displaystyle{ \iiint\limits_V { x^2 + y^2 + z^2 ~dV } }$$ where V is the unit ball $$x^2 + y^2 + z^2 \leq 1$$, using spherical coordinates.

Final Answer

$$4 \pi/5$$

Problem Statement

Evaluate $$\displaystyle{ \iiint\limits_V { x^2 + y^2 + z^2 ~dV } }$$ where V is the unit ball $$x^2 + y^2 + z^2 \leq 1$$, using spherical coordinates.

Solution

1935 video

video by PatrickJMT

Final Answer

$$4 \pi/5$$

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Integrate $$\displaystyle{ f(x,y,z) = 3e^{(x^2 + y^2 + z^2)^{3/2}} }$$ over the region inside $$x^2 + y^2 + z^2 = 1$$ in the first octant, using spherical coordinates.

Problem Statement

Integrate $$\displaystyle{ f(x,y,z) = 3e^{(x^2 + y^2 + z^2)^{3/2}} }$$ over the region inside $$x^2 + y^2 + z^2 = 1$$ in the first octant, using spherical coordinates.

Final Answer

$$\pi(e-1)/2$$

Problem Statement

Integrate $$\displaystyle{ f(x,y,z) = 3e^{(x^2 + y^2 + z^2)^{3/2}} }$$ over the region inside $$x^2 + y^2 + z^2 = 1$$ in the first octant, using spherical coordinates.

Solution

1936 video

video by MIP4U

Final Answer

$$\pi(e-1)/2$$

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Determine the volume outside $$x^2 + y^2 + z^2 = 1$$ and inside $$x^2 + y^2 + z^2 = 4$$.

Problem Statement

Determine the volume outside $$x^2 + y^2 + z^2 = 1$$ and inside $$x^2 + y^2 + z^2 = 4$$.

Final Answer

$$28 \pi/3$$

Problem Statement

Determine the volume outside $$x^2 + y^2 + z^2 = 1$$ and inside $$x^2 + y^2 + z^2 = 4$$.

Solution

1937 video

video by MIP4U

Final Answer

$$28 \pi/3$$

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Calculate the volume of the region outside the cone $$\phi = \pi/3$$ and inside the sphere $$\rho = 6 \cos \phi$$.

Problem Statement

Calculate the volume of the region outside the cone $$\phi = \pi/3$$ and inside the sphere $$\rho = 6 \cos \phi$$.

Final Answer

$$9 \pi/4$$

Problem Statement

Calculate the volume of the region outside the cone $$\phi = \pi/3$$ and inside the sphere $$\rho = 6 \cos \phi$$.

Solution

1938 video

video by MIP4U

Final Answer

$$9 \pi/4$$

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Use spherical coordinates to evaluate $$\iiint_V{ ( x^2 + y^2 + z^2 )^2 ~dV }$$ where V is the ball with center $$(0,0,0)$$ and radius 5.

Problem Statement

Use spherical coordinates to evaluate $$\iiint_V{ ( x^2 + y^2 + z^2 )^2 ~dV }$$ where V is the ball with center $$(0,0,0)$$ and radius 5.

Final Answer

$$312500\pi/7$$

Problem Statement

Use spherical coordinates to evaluate $$\iiint_V{ ( x^2 + y^2 + z^2 )^2 ~dV }$$ where V is the ball with center $$(0,0,0)$$ and radius 5.

Solution

Note: Although she calls this a volume in the video, it really is not a volume. Basically, she is evaluating the function $$( x^2 + y^2 + z^2 )^2$$ over the volume. If she was calculating the volume, the function would be 1.

1941 video

video by Krista King Math

Final Answer

$$312500\pi/7$$

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Evaluate, using spherical coordinates, $$\displaystyle{ \iiint_V{ \frac{1}{x^2+y^2+z^2}dV } }$$ where the volume V is between the spheres $$x^2 + y^2 + z^2 = 4$$ and $$x^2 + y^2 + z^2 = 25$$ in the first octant.

Problem Statement

Evaluate, using spherical coordinates, $$\displaystyle{ \iiint_V{ \frac{1}{x^2+y^2+z^2}dV } }$$ where the volume V is between the spheres $$x^2 + y^2 + z^2 = 4$$ and $$x^2 + y^2 + z^2 = 25$$ in the first octant.

Final Answer

$$3\pi/2$$

Problem Statement

Evaluate, using spherical coordinates, $$\displaystyle{ \iiint_V{ \frac{1}{x^2+y^2+z^2}dV } }$$ where the volume V is between the spheres $$x^2 + y^2 + z^2 = 4$$ and $$x^2 + y^2 + z^2 = 25$$ in the first octant.

Solution

1942 video

video by MIP4U

Final Answer

$$3\pi/2$$

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Calculate the volume of a sphere of radius R where $$\phi$$ ranges from 0 to $$\pi/6$$.

Problem Statement

Calculate the volume of a sphere of radius R where $$\phi$$ ranges from 0 to $$\pi/6$$.

Hint

The integral you set up should look like $$V = \iiint dV = \int_0^{\pi/6}{ \int_0^{2\pi}{ \int_0^R{ \rho^2\sin\phi ~ d\rho ~d\theta ~d\phi }}}$$.
Of course, since the limits of integration are constants, you can integrate in any order.

Problem Statement

Calculate the volume of a sphere of radius R where $$\phi$$ ranges from 0 to $$\pi/6$$.

Final Answer

$$\pi R^3 (2-\sqrt{3})/3$$

Problem Statement

Calculate the volume of a sphere of radius R where $$\phi$$ ranges from 0 to $$\pi/6$$.

Hint

The integral you set up should look like $$V = \iiint dV = \int_0^{\pi/6}{ \int_0^{2\pi}{ \int_0^R{ \rho^2\sin\phi ~ d\rho ~d\theta ~d\phi }}}$$.
Of course, since the limits of integration are constants, you can integrate in any order.

Solution

2224 video

video by Michel vanBiezen

Final Answer

$$\pi R^3 (2-\sqrt{3})/3$$

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Intermediate

Express as a triple integral, using spherical coordinates, the volume of the region above the cone $$z = \sqrt{x^2+y^2}$$ and inside the sphere $$x^2 + y^2 + z^2 = 2az, a > 0$$ and evaluate.

Problem Statement

Express as a triple integral, using spherical coordinates, the volume of the region above the cone $$z = \sqrt{x^2+y^2}$$ and inside the sphere $$x^2 + y^2 + z^2 = 2az, a > 0$$ and evaluate.

Final Answer

$$\pi a^3$$

Problem Statement

Express as a triple integral, using spherical coordinates, the volume of the region above the cone $$z = \sqrt{x^2+y^2}$$ and inside the sphere $$x^2 + y^2 + z^2 = 2az, a > 0$$ and evaluate.

Solution

Note - This same problem is set up in rectangular coordinates on the rectangular coordinates page and in cylindrical coordinates on the cylindrical coordinates page.

1933 video

video by Dr Chris Tisdell

Final Answer

$$\pi a^3$$

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Topics You Need To Understand For This Page

 spherical coordinates partial integrals triple integrals triple integrals in cylindrical coordinates

Trig Identities and Formulas

basic trig identities

$$\sin^2\theta+\cos^2\theta=1$$   |   $$1+\tan^2\theta=\sec^2\theta$$

$$\displaystyle{\tan\theta=\frac{\sin\theta}{\cos\theta}}$$   |   $$\displaystyle{\cot\theta=\frac{\cos\theta}{\sin\theta}}$$

$$\displaystyle{\sec\theta=\frac{1}{\cos\theta}}$$   |   $$\displaystyle{\csc\theta=\frac{1}{\sin\theta}}$$

power reduction (half-angle) formulae

$$\displaystyle{\sin^2\theta=\frac{1-\cos(2\theta)}{2}}$$   |   $$\displaystyle{\cos^2\theta=\frac{1+\cos(2\theta)}{2}}$$

double angle formulae

$$\sin(2\theta)=2\sin\theta\cos\theta$$   |   $$\cos(2\theta)=\cos^2\theta-\sin^2\theta$$

links

list of trigonometric identities - wikipedia

trig sheets - pauls online notes

17calculus trig formulas - full list

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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 Setting Up and Evaluating Triple Integrals Applications Practice

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