17Calculus - Triple Integrals in Spherical Coordinates
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.
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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 |
\(\iiint_V{f~dV}\) |
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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 |
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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 |
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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 |
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\( 4 \pi/5 \)
Problem Statement |
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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 |
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1935 video
video by PatrickJMT |
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Final Answer |
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\( 4 \pi/5 \) |
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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 |
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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.
Final Answer |
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\( \pi(e-1)/2 \)
Problem Statement |
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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.
Solution |
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1936 video
video by MIP4U |
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Final Answer |
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\( \pi(e-1)/2 \) |
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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 |
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Determine the volume outside \( x^2 + y^2 + z^2 = 1 \) and inside \( x^2 + y^2 + z^2 = 4 \).
Final Answer |
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\( 28 \pi/3 \)
Problem Statement |
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Determine the volume outside \( x^2 + y^2 + z^2 = 1 \) and inside \( x^2 + y^2 + z^2 = 4 \).
Solution |
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1937 video
video by MIP4U |
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Final Answer |
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\( 28 \pi/3 \) |
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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 |
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Calculate the volume of the region outside the cone \( \phi = \pi/3 \) and inside the sphere \( \rho = 6 \cos \phi \).
Final Answer |
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\( 9 \pi/4 \)
Problem Statement |
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Calculate the volume of the region outside the cone \( \phi = \pi/3 \) and inside the sphere \( \rho = 6 \cos \phi \).
Solution |
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1938 video
video by MIP4U |
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Final Answer |
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\( 9 \pi/4 \) |
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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 |
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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.
Final Answer |
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\( 312500\pi/7 \)
Problem Statement |
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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.
Solution |
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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 |
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Final Answer |
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\( 312500\pi/7 \) |
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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 |
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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.
Final Answer |
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\( 3\pi/2 \)
Problem Statement |
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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.
Solution |
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1942 video
video by MIP4U |
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Final Answer |
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\( 3\pi/2 \) |
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Calculate the volume of a sphere of radius R where \( \phi \) ranges from 0 to \( \pi/6 \).
Problem Statement |
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Calculate the volume of a sphere of radius R where \( \phi \) ranges from 0 to \( \pi/6 \).
Hint |
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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 |
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Calculate the volume of a sphere of radius R where \( \phi \) ranges from 0 to \( \pi/6 \).
Final Answer |
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\( \pi R^3 (2-\sqrt{3})/3 \)
Problem Statement |
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Calculate the volume of a sphere of radius R where \( \phi \) ranges from 0 to \( \pi/6 \).
Hint |
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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 |
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2224 video
video by Michel vanBiezen |
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Final Answer |
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\( \pi R^3 (2-\sqrt{3})/3 \) |
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Intermediate |
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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 |
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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 |
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\( \pi a^3 \)
Problem Statement |
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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 |
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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 |
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Final Answer |
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\( \pi a^3 \) |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Trig Identities and Formulas
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basic trig identities |
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\(\sin^2\theta+\cos^2\theta=1\) | \(1+\tan^2\theta=\sec^2\theta\) |
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\(\displaystyle{\tan\theta=\frac{\sin\theta}{\cos\theta}}\) | \(\displaystyle{\cot\theta=\frac{\cos\theta}{\sin\theta}}\) |
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\(\displaystyle{\sec\theta=\frac{1}{\cos\theta}}\) | \(\displaystyle{\csc\theta=\frac{1}{\sin\theta}}\) |
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power reduction (half-angle) formulae |
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\(\displaystyle{\sin^2\theta=\frac{1-\cos(2\theta)}{2}}\) | \(\displaystyle{\cos^2\theta=\frac{1+\cos(2\theta)}{2}}\) |
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double angle formulae |
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\(\sin(2\theta)=2\sin\theta\cos\theta\) | \(\cos(2\theta)=\cos^2\theta-\sin^2\theta\) |
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links |
Related Topics and Links
related topics |
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external links you may find helpful |
Pauls Online Notes: Triple Integrals in Spherical Coordinates |
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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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Practice Instructions
Unless otherwise instructed, evaluate these integrals in the spherical coordinate system.
