- #1
skanda9051
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Please some one help me how to solve this problem
integral-r^2 sin(theta) d(theta) d(phi)
integral-r^2 sin(theta) d(theta) d(phi)
Solve Surface Integral: r^2 sin(theta)
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SUMMARY
The discussion focuses on solving the surface integral of the expression r^2 sin(theta) d(theta) d(phi) in the context of magnetic flux E through a sphere of radius r. The integral is defined as integral E. da = integral (1/4pi Eo) (q/r^2) (r^2 sin(theta) d(theta) d(phi)). The participants clarify that this is a double integral with limits for both theta (0 to pi) and phi (0 to 2pi), which are determined by spherical coordinates. The integration process for both variables is essential for solving the problem accurately.
PREREQUISITES- Spherical coordinates and their limits
- Understanding of double integrals
- Basic knowledge of magnetic flux concepts
- Integration techniques for trigonometric functions
- Study the properties of spherical coordinates in calculus
- Learn about double integrals and their applications
- Explore magnetic flux calculations in electromagnetism
- Practice integrating trigonometric functions over specified limits
Students and professionals in physics and engineering, particularly those studying electromagnetism and calculus, will benefit from this discussion.
- #2
I like Serena
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MHB
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Hi skanda9051! 
What did you try?
That would help me to know what I should explain to you.
What did you try?
That would help me to know what I should explain to you.
- #3
skanda9051
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Well its magnetic flux E through a sphere of radius r and flux is given.
integral E. da=integral 1/4pi Eo (q/r^2).(r^2 sin(theta) d(theta) d(phi) they have given answer as q/Eo:-). My doubt is since it is surface integral there should be 2 limits one with respect to theta and another with respect to phi:-). So how did they integrate with respect to phi
integral E. da=integral 1/4pi Eo (q/r^2).(r^2 sin(theta) d(theta) d(phi) they have given answer as q/Eo:-). My doubt is since it is surface integral there should be 2 limits one with respect to theta and another with respect to phi:-). So how did they integrate with respect to phi
- #4
I like Serena
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Yes, there are limits wrt theta and another wrt to phi.
This is be a double integral and not a single integral.
The limits are defined by the definition of spherical coordinates, although you do not need them to integrate your expression.
How would you integrate \int 5 d\phi?
And how would you integrate \int 5 \sin(\theta) d\theta?
Btw, in (these) spherical coordinates phi runs from 0 to 2pi, and theta runs from 0 to pi.
This is be a double integral and not a single integral.
The limits are defined by the definition of spherical coordinates, although you do not need them to integrate your expression.
How would you integrate \int 5 d\phi?
And how would you integrate \int 5 \sin(\theta) d\theta?
Btw, in (these) spherical coordinates phi runs from 0 to 2pi, and theta runs from 0 to pi.
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