# Spherical surface area element

• binbagsss
So what is the problem you are trying to solve?In summary, the expression Rdθ.2∏Rsinθ represents the width and length of a strip on a sphere with radius R, where θ and φ are the variables being integrated over. The surface area of this strip can be found by multiplying these two values and integrating over the sphere. This can be used to find the total charge and electric field according to Coulomb's law. However, this calculation may vary depending on the given surface density and the angle between the surface normal and the radial component of the constant vector P.
binbagsss
See image attached.

I am trying to understand the expression : Rdθ.2∏Rsinθ

Here are my thoughts so far:

Rdθ is the width of a strip, θ being the variable changing/to integrate over, giving arise to the elements.

2∏Rsinθ must then be the length. However I don't understand this expression.

I think I understand Rdθ - it is the arc length.

Here are my length thoughts: From the triangle sin θ =(l/2)/R, where l is the length we are after, and l/2, because this triangle only gives half of the length we are after.

Ofc this must be wrong, as I have a ∏ unexplained.

Many thanks to anyone who can help shed some light on this, greatly appreciated !

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binbagsss said:
See image attached.

I am trying to understand the expression : Rdθ.2∏Rsinθ
Here are my thoughts so far:

Rdθ is the width of a strip, θ being the variable changing/to integrate over, giving arise to the elements.

2∏Rsinθ must then be the length. However I don't understand this expression.

I think I understand Rdθ - it is the arc length.

Here are my length thoughts: From the triangle sin θ =(l/2)/R, where l is the length we are after, and l/2, because this triangle only gives half of the length we are after.

Ofc this must be wrong, as I have a ∏ unexplained.Many thanks to anyone who can help shed some light on this, greatly appreciated !

It would help if you give us a statement of what problem is being solved. Also, in the figure it doesn't appear that R is the radius of the circle. Or maybe it is, just overwritten on the vertical strip.

I am given a surface change densiity σ = P.$\hat{n}$ = P cos θ, where P is a constant, and am multiplying this by the area element to find the total charge.
(I am then going to apply Coulomb's law to find the total electric field).

Does this help?

My apologies, R is the radius of the circle

It doesn't help me. I don't see what you are doing. I don't see any spherical dS element and I don't understand the problem. Maybe someone else will stop by.

anyone?

binbagsss said:
See image attached.

I am trying to understand the expression : Rdθ.2∏Rsinθ
Here are my thoughts so far:

Rdθ is the width of a strip, θ being the variable changing/to integrate over, giving arise to the elements.

2∏Rsinθ must then be the length. However I don't understand this expression.

I think I understand Rdθ - it is the arc length.

$r\,d\theta$ is the arclength in the direction of increasing $\theta$ (with $\phi$ and $r$ fixed). Arclength in the direction of increasing $\phi$ (with $\theta$ and $r$ fixed) is $r\sin\theta\,d\phi$. The area of the surface corresponding to $[\theta, \theta + \delta\theta]\times[\phi,\phi + \delta\phi]$ with $r = R$ constant is then approximately $(R\delta\theta)(R \sin (\theta)\delta\phi) = R^2 \sin (\theta)\delta\theta\delta\phi$. Thus the integral of $\sigma$ over the sphere $r = R$ is
$$\int_{r = R} \sigma\,dS = \int_0^{2\pi} \int_0^{\pi} \sigma R^2 \sin \theta\,d\theta\,d\phi$$ and if $\sigma$ is independent of $\phi$ this reduces to
$$R^2\left(\int_0^{2\pi}\,d\phi\right)\int_0^\pi \sigma\sin \theta\,d\theta = 2\pi R^2 \int_0^\pi \sigma\sin\theta\,d\theta.$$

binbagsss said:
I am given a surface change densiity σ = P.$\hat{n}$ = P cos θ, where P is a constant,

Are you given that $\sigma = P \cos \theta$ for constant $P$, or are you given $\sigma = \mathbf{P} \cdot \hat{\mathbf{n}}$ for constant $\mathbf{P}$? I can't tell whether $\mathbf{P} \cdot \hat{\mathbf{n}} = P\cos\theta$ is something you were given or the reflex action of calling the angle in the definition of the dot product "$\theta$". But that angle might not be equal to the coordinate $\theta$, so should be given a different name.

Assuming $\hat{\mathbf{n}}$ is the outward normal to the sphere, you have $\hat{\mathbf{n}} = \hat{\mathbf{r}}$, so $\mathbf{P} \cdot \hat{\mathbf{n}}$ is the radial component of $\mathbf{P}$. But if $\mathbf{P}$ is constant then its radial component is not constant but depends on both $\theta$ and $\phi$.

Last edited:

## What is a spherical surface area element?

A spherical surface area element is a small piece or section of a spherical surface. It is used in calculations to determine the total surface area of a sphere or to calculate flux across a spherical surface.

## How is the spherical surface area element calculated?

The spherical surface area element is calculated using the formula dA = r²sinθdθdφ, where r is the radius of the sphere and θ and φ are the angles that define the position of the element on the surface.

## What is the importance of the spherical surface area element in physics?

The spherical surface area element is important in various fields of physics, including electromagnetism, thermodynamics, and fluid mechanics. It allows for the calculation of surface integrals and flux across spherical surfaces, which are key concepts in these fields.

## How does the spherical surface area element relate to the surface area of a sphere?

The spherical surface area element represents the differential change in surface area of a sphere. By integrating this element over a spherical surface, we can calculate the total surface area of the sphere.

## Can the spherical surface area element be used for non-spherical surfaces?

No, the spherical surface area element is specifically designed for spherical surfaces. For non-spherical surfaces, other surface area elements such as the cylindrical surface area element or the planar surface area element should be used.

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