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 

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 

\( \pi(e1)/2 \)
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 

video by MIP4U 

Final Answer 

\( \pi(e1)/2 \) 
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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 

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 

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.
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 

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 

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.
video by Dr Chris Tisdell 

Final Answer 

\( \pi a^3 \) 
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You CAN Ace Calculus
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 (halfangle) 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 
related topics 

external links you may find helpful 
Pauls Online Notes: Triple Integrals in Spherical Coordinates 
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Practice Instructions
Unless otherwise instructed, evaluate these integrals in the spherical coordinate system.