Integral (area) of a revolved parabola

[Karol]

Whats's the volume under the revolving parabola y=ax². R is x_max.

$$V=\int_0^R\pi xdx\cdot y=\pi\int_o^R x\cdot Ax^2dx=\pi A\frac{R^4}{4}$$
Volume should be relative to R³. and if i had, for example, y=Ax³ then, according to my calculation i would get relative to R⁷ and so on.

 

[Samy_A]

Volume should be relative to R³.

Indeed (proportional would probably be a better term).

and if i had, for example, y=Ax³ then, according to my calculation i would get relative to R⁷ and so on.

So clearly something is wrong here.

What formula are you using to find the volume of a solid of revolution?

 

[Ssnow]

Why there is $x\cdot y$ in your formula, this must be $V=\int_{0}^{R}\pi f^{2}(x)\,d\,x$...

 

[Karol]

The volume is the sum of the vertical tubes of changing height y and infinitesimally small thickness dx. the perimeter of the tubes is πr.
$$V=\int dv=\int (\pi x)y\cdot dx$$
y, in turn, equals Ax². but according to this approach i would get a bigger error if i had, for example, y=Bx³

 

[muscaria]

Volume should be relative to R3

There is no reason for the volume to be proportional to $R^3$. In fact you will discover that it isn't!

The volume is the sum of the vertical tubes of changing height y and infinitesimally small thickness dx. the perimeter of the tubes is πr.
$$V=\int dv=\int (\pi x)y\cdot dx$$
y, in turn, equals Ax². but according to this approach i would get a bigger error if i had, for example, y=Bx³

I'm not sure why you are considering the perimeter of the disks centred about the x-axis. Would you not want to be using the area of the disks and then integrating over x? If you revolve your parabola about the x-axis, you get lots of solid disks centred about the x-axis which grow in radius as you move further along in x, right? What is the radius of a given disk as a function of x? Once you have this, you then know the area of each disk as a function of x. Once you have an expression for the disk area as a function of x, you can integrate over x, effectively summing up the infinitesimal cylinders centred about the x-axis. Does this make sense?

 

[Karol]

Would you not want to be using the area of the disks

$$V=\int_0^{y=AR^2}\pi(R^2-x^2)dy=\pi\int_0^{AR^2}R^2-\frac{y}{A}dy=\pi\left[ R^2y-\frac{y^2}{2A} \right]_0^{AR^2}=\frac{\pi A}{2}R^4$$
Also with my previous method of tubes:
$$V=\int_0^R2\pi xdx\cdot y=2\pi\int_o^R x\cdot Ax^2dx=\frac{\pi A}{2}R^4$$
I found in a site the volume under a paraboloid, an inverse of mine, is $V=\frac{\pi}{2}hr^2$, and according to that my volume is:
$$V=\pi R^2h-\frac{\pi}{2}hR^2=...=\frac{\pi A}{2}R^4$$
In all 3 methods i get the same.
But the units are $R^4[m^4]$ while volume is [m³]

 

[Samy_A]

$$V=\int_0^{y=AR^2}\pi(R^2-x^2)dy=\pi\int_0^{AR^2}R^2-\frac{y}{A}dy=\pi\left[ R^2y-\frac{y^2}{2A} \right]_0^{AR^2}=\frac{\pi A}{2}R^4$$
Also with my previous method of tubes:
$$V=\int_0^R2\pi xdx\cdot y=2\pi\int_o^R x\cdot Ax^2dx=\frac{\pi A}{2}R^4$$
I found in a site the volume under a paraboloid, an inverse of mine, is $V=\frac{\pi}{2}hr^2$, and according to that my volume is:
$$V=\pi R^2h-\frac{\pi}{2}hR^2=...=\frac{\pi A}{2}R^4$$
In all 3 methods i get the same.
But the units are $R^4[m^4]$ while volume is [m³]

I made the same mistake as you. In what units is A?

 

[Svein]

Whats's the volume under the revolving parabola y=ax². R is x_max.
View attachment 92398
$$V=\int_0^R\pi xdx\cdot y=\pi\int_o^R x\cdot Ax^2dx=\pi A\frac{R^4}{4}$$
Volume should be relative to R³. and if i had, for example, y=Ax³ then, according to my calculation i would get relative to R⁷ and so on.

Yes. Your trouble lies with the interpretation of the constant A. Since you are trying to find a volume, you are assigning a "length dimension" to x. Assume that x has the dimension m (meter). Then y also needs to have the dimension m. Using the formula y = Ax² means that A must have the dimension m^-1. With that in mind the dimension of your volume becomes m^-1⋅m⁴ = m³.

 

[Karol]

I understood, thanks