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Magazine Chapter 12: Problem 23
\(21-32\) Use spherical coordinates. Evaluate \(\iiint_{E}\left(x^{2}+y^{2}\right) d V,\) where \(E\) lies between the spheres \(x^{2}+y^{2}+z^{2}=4\) and \(x^{2}+y^{2}+z^{2}=9\)
Short Answer
Expert verified
The value of the integral is \(\frac{1688\pi}{15}\).
Step by step solution
01
Convert Cartesian to Spherical Coordinates
The volume element in spherical coordinates is given by \[ dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]. \For the integrand \(x^2 + y^2\), we note that \(x^2 + y^2 = \rho^2 \sin^2 \phi\). We'll integrate over \(\theta\) from 0 to \(2\pi\), over \(\phi\) from 0 to \(\pi\), and \(\rho\) between the radii given by the spheres, which is from 2 to 3.
02
Set Up the Triple Integral
The integral in spherical coordinates becomes \[\int_0^{2\pi} \int_0^{\pi} \int_2^3 \rho^2 \sin^2 \phi \cdot \rho^2 \sin \phi \, d \rho \, d \phi \, d \theta \]. Simplify the integrand to \[ \rho^4 \sin^3 \phi \].
03
Integrate with Respect to \(\rho\)
Evaluate the inner integral with respect to \(\rho\): \[\int_2^3 \rho^4 \, d \rho = \left[ \frac{\rho^5}{5} \right]_2^3 = \frac{3^5}{5} - \frac{2^5}{5} = \frac{243}{5} - \frac{32}{5} = \frac{211}{5} \].
04
Integrate with Respect to \(\phi\)
Now integrate with respect to \(\phi\):\[\int_0^{\pi} \sin^3 \phi \, d \phi \].Use the identity \(\sin^3 \phi = (1 - \cos^2 \phi) \sin \phi\) and the substitution \(u = \cos \phi\), giving \(du = -\sin \phi \, d \phi\), so the integral becomes \[\int_1^{-1} (1 - u^2) (-du) = \int_{-1}^{1} (1 - u^2) \, du \].This evaluates to \[\left[ u - \frac{u^3}{3} \right]_{-1}^{1} = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}\].
05
Integrate with Respect to \(\theta\)
The outermost integral is with respect to \(\theta\), a simple integral:\[\int_0^{2\pi} \, d \theta = 2\pi\].
06
Combine the Results
Putting it all together, the final value of the integral is:\[2\pi \times \frac{4}{3} \times \frac{211}{5} = \frac{1688\pi}{15}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Triple Integrals
A triple integral is used to compute volumes and other quantities in three-dimensional space. It is an extension of double integrals, which you might already be familiar with from calculating areas in two-dimensional regions. When dealing with functions of three variables, triple integrals become especially handy.
In our exercise, the volume of a certain region in space is determined by integrating a function over that region using spherical coordinates. When performing a triple integral, we typically iterate through three nested integrals, each corresponding to a different variable.
The limits of each integral specify the span over which the variable will vary. Spherical coordinates are often used in triple integrals, especially when the region is a sphere or a spherical shell, as it simplifies the integration process.
In our exercise, the volume of a certain region in space is determined by integrating a function over that region using spherical coordinates. When performing a triple integral, we typically iterate through three nested integrals, each corresponding to a different variable.
The limits of each integral specify the span over which the variable will vary. Spherical coordinates are often used in triple integrals, especially when the region is a sphere or a spherical shell, as it simplifies the integration process.
Volume Element
When switching to spherical coordinates, the volume element \(dV\) becomes \(\rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\). This expression accounts for the geometry of spherical coordinates. It takes into account the facts that:
This calculation, accounting for the spherical volume element, is a key part of integrating in spherical coordinates. Integrating \(x^2+y^2\) essentially transforms into \(\rho^2\sin^2\phi\) due to the conversion from Cartesian to spherical coordinates.
- The radial distance \(\rho\) is cubed for volume considerations in addition to one more factor from the spherical measure.
- The angle \(\phi\) from the positive z-axis is adjusted by a factor of \(\sin \phi\) due to its relation to the radius of the horizontal section.
- The azimuthal angle \(\theta\) ranges from \(0\) to \(2\pi\), covering the full range around the z-axis.
This calculation, accounting for the spherical volume element, is a key part of integrating in spherical coordinates. Integrating \(x^2+y^2\) essentially transforms into \(\rho^2\sin^2\phi\) due to the conversion from Cartesian to spherical coordinates.
Cartesian to Spherical Conversion
Converting from Cartesian to spherical coordinates is a crucial step in solving integrals over spherical regions. This conversion makes the integration much more manageable.
Here's how it goes:
Spherical coordinates use the radius \(\rho\) from the origin to the point, the angle \(\theta\) around the z-axis, and the angle \(\phi\) from the z-axis. The integrand \(x^2 + y^2\) converts to \(\rho^2\sin^2\phi\), simplifying integration over a spherical region and capitalizing on the symmetrical properties of spheres.
This transformation is essential when dealing with boundaries determined by spheres, as seen in our regions defined by \(x^2+y^2+z^2=4\) and \(x^2+y^2+z^2=9\).
Here's how it goes:
- \(x = \rho \sin \phi \cos \theta\)
- \(y = \rho \sin \phi \sin \theta\)
- \(z = \rho \cos \phi\)
Spherical coordinates use the radius \(\rho\) from the origin to the point, the angle \(\theta\) around the z-axis, and the angle \(\phi\) from the z-axis. The integrand \(x^2 + y^2\) converts to \(\rho^2\sin^2\phi\), simplifying integration over a spherical region and capitalizing on the symmetrical properties of spheres.
This transformation is essential when dealing with boundaries determined by spheres, as seen in our regions defined by \(x^2+y^2+z^2=4\) and \(x^2+y^2+z^2=9\).
