- #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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In summary, the conversation discusses solving a problem involving a surface integral of a magnetic flux through a sphere. The person is confused about the limits of integration and how to integrate with respect to phi. It is clarified that there are two limits, one for theta and one for phi, and that the limits are defined by spherical coordinates. The conversation then gives examples of integrating with respect to phi and theta in spherical coordinates.
- #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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MHB
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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 [itex]\int 5 d\phi[/itex]?
And how would you integrate [itex]\int 5 \sin(\theta) d\theta[/itex]?
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 [itex]\int 5 d\phi[/itex]?
And how would you integrate [itex]\int 5 \sin(\theta) d\theta[/itex]?
Btw, in (these) spherical coordinates phi runs from 0 to 2pi, and theta runs from 0 to pi.
1. What is the meaning of "r^2 sin(theta)" in the surface integral?
The term "r^2 sin(theta)" represents the geometric relationship between the distance from the origin to a point on a surface (r) and the angle between the normal vector to the surface and the positive z-axis (theta). It is used to calculate the surface area element in polar coordinates.
2. How is the surface integral of "r^2 sin(theta)" calculated?
The surface integral of "r^2 sin(theta)" is calculated by first breaking down the surface into small patches or elements, calculating the area of each element using the formula A = r^2 sin(theta) dA, and then taking the sum of all the areas to get the total surface integral.
3. Can "r^2 sin(theta)" be used for any type of surface?
Yes, "r^2 sin(theta)" can be used for any type of surface that can be described in polar coordinates. This includes surfaces such as spheres, cones, and cylinders.
4. How does the value of "r^2 sin(theta)" affect the surface integral?
The value of "r^2 sin(theta)" affects the surface integral by determining the size of the surface area element. A larger value of "r^2 sin(theta)" means a larger surface area element, and thus a larger overall surface integral.
5. Are there any real-life applications of "r^2 sin(theta)" in surface integrals?
Yes, "r^2 sin(theta)" has various real-life applications in fields such as physics, engineering, and mathematics. For example, it can be used to calculate the moment of inertia of a rotating object, the electric flux through a curved surface, and the force exerted by a fluid on a submerged object.
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