Evaluate the spherical coordinate integrals

In summary, the spherical coordinate integrals can be evaluated using the following steps:1. Integrate the function over the sphere using the spherical law of cosines.2. Use the divergence theorem to calculate the second integral.3. Evaluate the first integral using the divergence theorem.
  • #1
karush
Gold Member
MHB
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$\textsf{Evaluate the spherical coordinate integrals}$
\begin{align*}\displaystyle
DV_{22}&=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{2}
\, (\rho \cos \phi) \rho^2 \sin \phi
\, d\rho \, d\phi \, d\theta \\
%&=\color{red}{abc}
\end{align*}
so then next ?
\begin{align*}\displaystyle
DV_{22}&=2\pi\int_{0}^{\pi/4}\int_{0}^{2}
\, (\rho \cos \phi) \rho^2 \sin \phi
\, d\rho \, d\phi \,
\end{align*}

no book answer:mad:
 
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  • #2
\(\displaystyle DV_{22}= 2\pi \left(\int_0^{\pi/4} sin(\phi)cos(\phi) d\phi\right)\left(\int_0^2 \rho^3 d\rho\right)\)

Both of those integrals should be easy.
 
  • #3
Country Boy said:
\(\displaystyle DV_{22}= 2\pi \left(\int_0^{\pi/4} sin(\phi)cos(\phi) d\phi\right)\left(\int_0^2 \rho^3 d\rho\right)\)

Both of those integrals should be easy.

are you suggesting multiplication

well here is some of it\begin{align*}
&\displaystyle\int_{0}^{2}\rho^3 \, dp
=\left[\frac{\rho^4}{4} \right]_{0}^{2}
=4
\end{align*}

\begin{align*}\displaystyle
&\int_0^{\pi/4} \sin(\phi)\cos(\phi) \, d\phi\\
=&\frac{1}{2}\int_0^{\pi/4}
2\sin(\phi)\cos(\phi)\, d\phi\\
=&\frac{1}{2}\int_0^{\pi/4}
\sin(2\phi)\, d\phi\\
=&\frac{1}{2}\biggr[2\cos(2\phi) \biggr]_0^{\pi/4}\\
=&4
\end{align*}
 
Last edited:
  • #4
karush said:
$\textsf{Evaluate the spherical coordinate integrals}$
\begin{align*}\displaystyle
DV_{22}&=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{2}
\, (\rho \cos \phi) \rho^2 \sin \phi
\, d\rho \, d\phi \, d\theta \\
%&=\color{red}{abc}
\end{align*}
so then next ?
\begin{align*}\displaystyle
DV_{22}&=2\pi\int_{0}^{\pi/4}\int_{0}^{2}
\, (\rho \cos \phi) \rho^2 \sin \phi
\, d\rho \, d\phi \,
\end{align*}

no book answer:mad:

I would begin by writing:

\(\displaystyle I=\int_0^{2\pi}\int_0^{\Large\frac{\pi}{4}} \sin(\phi)\cos(\phi)\int_0^2 \rho^3\,d\rho\,d\phi\,d\theta\)

Next, work your way from the inside out:

\(\displaystyle I=4\int_0^{2\pi}\int_0^{\Large\frac{\pi}{4}} \sin(\phi)\cos(\phi)\,d\phi\,d\theta=4\int_0^{2\pi}\int_0^{\Large\frac{1}{\sqrt{2}}} u\,du\,d\theta\)

\(\displaystyle I=\int_0^{2\pi}\,d\theta=2\pi\)
 
  • #5
where does the \(\displaystyle 4\) come from?what is u ?
 
  • #6
karush said:
where does the \(\displaystyle 4\) come from?

It came from the evaluation of the innermost integral.

karush said:
what is u ?

\(\displaystyle u=\sin(\phi)\implies du=\cos(\phi)\,d\phi\)
 
  • #7
MarkFL said:
It came from the evaluation of the innermost integral.
\(\displaystyle u=\sin(\phi)\implies du=\cos(\phi)\,d\phi\)

so then we have

$$8\pi\int_{0}^{\sqrt{2}/2} u\,du$$
 
  • #8
karush said:
so then we have

$$8\pi\int_{0}^{\sqrt{2}/2} u\,du$$

You could look at it that way, but I prefer to work from the inside out. :)
 
  • #9
like this?

\(\displaystyle

\begin{align*}\displaystyle &4\int_0^{2\pi}\int_0^{\Large\frac{1}{\sqrt{2}}} u\,du\,d\theta\\
=&4\int_0^{2\pi}
\left[\frac{u^2}{2} \right]_0^{\sqrt{2}/2}
d\theta
\end{align*}\)
 

FAQ: Evaluate the spherical coordinate integrals

1. What are spherical coordinates?

Spherical coordinates are a system of coordinates used to locate points in three-dimensional space. They are defined by a radius, an azimuth angle, and an elevation angle.

2. How do you convert spherical coordinates to Cartesian coordinates?

To convert spherical coordinates to Cartesian coordinates, you can use the following formulas:
x = r * sin(θ) * cos(ϕ)
y = r * sin(θ) * sin(ϕ)
z = r * cos(θ)
Where r is the radius, θ is the azimuth angle, and ϕ is the elevation angle.

3. What is the difference between a triple integral and a spherical coordinate integral?

A triple integral is a type of integral used to calculate the volume under a three-dimensional surface. A spherical coordinate integral is a specific type of triple integral that is used when the integrand is expressed in terms of spherical coordinates.

4. What are some applications of spherical coordinate integrals?

Spherical coordinate integrals are commonly used in physics and engineering to solve problems involving three-dimensional objects, such as calculating the mass, center of mass, and moment of inertia of a solid object. They are also used in electromagnetism to calculate electric and magnetic fields.

5. How do you evaluate a spherical coordinate integral?

To evaluate a spherical coordinate integral, you can use the following steps:
1. Convert the integrand from Cartesian coordinates to spherical coordinates using the appropriate substitution.
2. Determine the limits of integration for each variable (radius, azimuth angle, elevation angle).
3. Use the appropriate integration techniques (such as u-substitution or trigonometric identities) to solve the integral.
4. Convert the resulting integral back to Cartesian coordinates if necessary.

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