# Arclength, Surface Area and Volume

• Eidos
In summary, if you want the arclength of a function, you need to define the function explicitly and not implicitly.
Eidos
I was just thinking:

If $$\iint dS$$ is the surface area of a level surface, S, and $$\iiint dV$$ is the volume of an enclosed solid, V, shouldn't $$\int df$$ be the arclength of a function f(x)?

Lets say that our surface is given implicitly by $$\Phi$$
For the surface area we get:
$$\iint dS$$ = $$\int_{y_0}^{y_1}\int_{x_0}^{x_1}|\nabla\Phi|dxdy$$
where $$|\nabla\Phi|$$ is the Jacobian determinant.

Now if we define function f implicitly by $$\alpha$$
$$\int df$$ = $$\int_{x_0}^{x_1} |\nabla \alpha| dx$$

Arclength is usually given by $$\int_{x_0}^{x_1} \sqrt{1+(\frac{dy}{dx})^2} dx$$.

This works out the same;
say our function is y=f(x) then $$\alpha=y-f(x)$$,
$$\nabla\alpha=(-f'(x),1,0)$$ whose modulus is exactly $$\sqrt{1+(\frac{dy}{dx})^2}$$

Is this correct?

I was just thinking:

If $$\iint dS$$ is the surface area of a level surface, S, and $$\iiint dV$$ is the volume of an enclosed solid, V, shouldn't $$\int df$$ be the arclength of a function f(x)?
Why should it? A surface has an area and a solid has a volume: both are geometric figures. A function is not a geometric figure and does not have an "arclength". If you meant "shouldn't $\int d\sigma$ be the arclength of the curve $\sigma$", then the answer is "yes, of course".

Lets say that our surface is given implicitly by $$\Phi$$
For the surface area we get:
$$\iint dS$$ = $$\int_{y_0}^{y_1}\int_{x_0}^{x_1}|\nabla\Phi|dxdy$$
where $$|\nabla\Phi|$$ is the Jacobian determinant.

Now if we define function f implicitly by $$\alpha$$
$$\int df$$ = $$\int_{x_0}^{x_1} |\nabla \alpha| dx$$

Arclength is usually given by $$\int_{x_0}^{x_1} \sqrt{1+(\frac{dy}{dx})^2} dx$$.

This works out the same;
say our function is y=f(x) then $$\alpha=y-f(x)$$,
$$\nabla\alpha=(-f'(x),1,0)$$ whose modulus is exactly $$\sqrt{1+(\frac{dy}{dx})^2}$$

Is this correct?
?? if y= f(x), then $\alpha= y- f(x)$ is identically 0 and so $\nabla\alpha= 0$.

Sorry about my fumble, $$\alpha$$ is an implicit function for the curve, similar to how $$\Phi$$ is an implicit function for a surface. We let $$\alpha$$ or $$\Phi$$ be equal to a constant and it kicks out a curve or surface respectively.

I was using a process analogous to:
z=z(x,y)
let $$\Phi=z-z(x,y)$$ so $$|\nabla \Phi|=\sqrt{1+z_x^2+z_y^2}$$ which is the jacobian determinant for an explicit surface. I do see your point regarding the fact that $$\Phi$$ is identically zero.

I should have called f a curve in the beginning, I must learn to be more precise.

Edit: Indeed I have put the cart before the horse, the trick is to define $$\Phi=z(x,y)-z$$ so that $$\Phi=0$$ is the level surface z=z(x,y). The rest follows from there. Similarly for the implicit curve $$\alpha$$. Thank you for pointing out my error

Last edited:

## 1. What is the formula for calculating arclength?

The formula for calculating arclength is: L = rθ, where L is the arclength, r is the radius of the circle, and θ is the central angle in radians.

## 2. How do you find the surface area of a three-dimensional shape?

The formula for finding the surface area of a three-dimensional shape depends on the specific shape. Some common formulas include: - Cube: 6s^2, where s is the length of one side- Cylinder: 2πrh + 2πr^2, where r is the radius and h is the height- Sphere: 4πr^2, where r is the radius- Cone: πr(r + √(h^2 + r^2)), where r is the radius and h is the height

## 3. Can you explain the concept of volume?

Volume is the measure of the amount of space occupied by a three-dimensional object. It is typically measured in cubic units (e.g. cubic inches, cubic centimeters). The formula for calculating volume also depends on the shape, but some common formulas include: - Cube: s^3, where s is the length of one side- Cylinder: πr^2h, where r is the radius and h is the height- Sphere: (4/3)πr^3, where r is the radius- Cone: (1/3)πr^2h, where r is the radius and h is the height

## 4. How does changing the radius affect the arclength and surface area of a circle?

As the radius of a circle increases, the arclength also increases. This is because a larger radius means a larger circumference, and therefore a longer arclength. On the other hand, changing the radius does not directly affect the surface area of a circle. The surface area only changes if the radius is used to calculate the surface area of a three-dimensional shape, such as a cylinder or sphere.

## 5. Can you provide an example of how to use these concepts in real life?

One example of using these concepts in real life is when designing a container for a certain amount of liquid. By using the formula for volume, you can calculate the necessary dimensions for the container to hold the desired amount of liquid. Additionally, knowing the surface area can be useful for determining the amount of material needed to construct the container. Another example is in construction, where arclength and volume calculations are used to determine the amount of materials needed for curved structures and the volume of concrete needed for foundations.

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