The calculation of the surface of a circle
The surface of the circle of radius \( R\) is worth:
$$ S_{circle} = \pi R^2 $$
The calculation of the surface of a sphere
The surface of a sphere of radius \( R\) is worth:
$$ S_{sphere} = 4\pi R^2 $$
The calculation of the surface of a cone
The cone is a pyramid with a circular base. It is characterized by its height \( h\) and the radius \( R\) from its base.
The surface of a cone of height \( h\) with a circular base of radius \( r\) is worth:
$$ S_{cone} = \pi r^2 + \pi r \sqrt{r^2 + h^2} $$
The calculation of the volume of a sphere
The volume of a sphere of radius \( R\) is worth:
$$ V_{sphere} = \frac{4}{3}\pi R^3 $$
The calculation of the volume of a cone
The volume of a cone of height \( h\) with a circular base of radius \( r\) is worth:
$$ V_{cone} = \frac{\pi r^2 h}{3} $$
Let be a circle of radius \( R\), and \( \theta \) an infinitesimal angle.
Let us now consider an infinitesimal isosceles triangle \( \Delta \) trained by \( \bigl\{x, R, Rd\theta \bigr\}\), and such as the following figure.
Let \( S_{\Delta} \) be the surface of this triangle. Being said that \( \Delta \) is infinitesimal, we consider that is right-angled between \( x \) and \( Rd\theta \).
$$ dS_{\Delta} = \frac{1}{2}R^2d\theta$$
We will then integrate the surface of \( \Delta \) on the curvilinear distance \( Rd\theta\). It is therefore necessary to add the different surfaces \(dS_{\Delta}\) on the angle \(\alpha\).
$$ \int_0^{\alpha} dS_{\Delta} = \int_0^{\alpha} \frac{1}{2}R^2d\theta $$
$$ S(\alpha) = \frac{1}{2}R^2 \int_0^{\alpha} d\theta $$
$$ S(\alpha) = \frac{1}{2}R^2 \Bigl[ \theta \Bigr]_0^{\alpha} $$
$$ S(\alpha) = \frac{1}{2}R^2\alpha$$
Now for \( \alpha = 2\pi \):
$$ S(2\pi) = \frac{2\pi}{2}R^2 $$
And finally,
$$ S_{circle} = \pi R^2 $$
Let us consider a circle of radius \( R\), and \( \theta \) an angle formed by the radius \( R\) and the base \( \overrightarrow{Ox}\).
As well, let us consider on infinitesimal curvilinear length \( dl = dR \theta\), such as the following figure:
In this figure, we have also added the mobile radius \(x\) depending on \(\theta\).
With the rules of trigonometry, we do have:
$$ \Biggl \{ \begin{align*} x = R cos(\theta) \qquad (1) \\ y = R sin(\theta) \end{align*}$$
By considering a moving perimeter of the circle as a function of \(x\), we do have:
$$ P(x) = 2\pi x \qquad (2) $$
By injecting \((1)\) into \((2)\):
$$ P(\theta) = 2\pi R cos(\theta) \qquad (3)$$
We then have an infinitesimal surface \( dS\), shown in red in the figure below.
$$ dS = P(\theta).dl \qquad (4) $$
But, we also have:
$$ dl = dR \theta \qquad (5) $$
Now, by injecting \((3)\) and \((5)\) into \((4)\):
$$ dS = 2\pi R cos(\theta) .dR \theta $$
We will integrate the angle \( \theta \) on \( [0, \frac{\pi}{2}] \), which will cover the surface of half a sphere.
$$ \int_0^{\frac{\pi}{2}} dS = \int_0^{\frac{\pi}{2}}2\pi R cos(\theta) .dR \theta $$
$$S_{\frac{1}{2}} = 2\pi R^2 \int_0^{\frac{\pi}{2}} cos(\theta).d \theta $$
$$S_{\frac{1}{2}} = 2\pi R^2\Bigl[ sin(\theta) \Bigr]_0^{\frac{\pi}{2}} $$
$$S_{\frac{1}{2}} = 2\pi R^2\Bigl( sin\left( \frac{\pi}{2}\right) - sin(0) \Bigr) $$
$$S_{\frac{1}{2}} = 2\pi R^2 $$
To have the complete surface, we will simply double this value.
$$ S_{sphere} = 2S_{\frac{1}{2}} $$
And as a result,
$$ S_{sphere} = 4\pi R^2 $$
Let us consider a cone of height \(h\) with a circular base of raidus \( R\).
For simplicity of demonstration, we have turned this cone upside down.
Let us call the surface of the base \(S_{base}\)and the surface around the vertical base \(S_{axis}\).
The calculation of the total surface of this cone is then:
$$ S_{cone} = S_{base} + S_{axis} $$
The surface of its base is the surface of a circle of radius \(r\), so:
$$ S_{base} = \pi r ^2 $$
In the following figure, we have added a movable height \(y\), as well as a mobile radius \(x\).
The moving radius \(x\) then varies depending on \(y\).
Let us note \(P(x)\) the moving perimeter of the circle, which is a function of \(x\). We do have:
$$ P(x) = 2\pi x \qquad (1) $$
By applying the Thales' theorem, we do have:
$$ \frac{x}{r} = \frac{y}{h} \Longleftrightarrow x = \frac{yr}{h} \qquad (2) $$
Therefore injecting \((2)\) into \((1)\), we obtain a mobile perimeter, this time according to \(y\) :
$$ P(y) = \frac{2 \pi yr}{h} $$
In order to cover the entire vertical surface, we will integrate this perimeter over the entire distance \(l\), hypotenuse of the right-angled triangle \(\{x, y, l\}\).
Let us now note \(S_{axis}(y)\) this surface to be calculated, which is a function of the mobile height \(y\).
And let us call \(dS\) an infinitesimal surface element, we do have:
$$ dS= P(y) \ dl $$
$$ \int_0^{h} dS = \int_0^{h} \frac{2 \pi yr}{h} \ dl $$
$$ \int_0^{h} dS = \frac{2 \pi r}{h} \int_0^{h} y \ dl \qquad (3) $$
Now, by applying the Pythagorean theorem, we notice that:
$$ l^2 = x^2 + y^2 $$
$$ l = \sqrt{x^2 + y^2} $$
Thus, if we want to keep an infinitesimal part \(dl\):
$$ dl = d\left(\sqrt{x^2 + y^2}\right) \qquad (4) $$
But, we previously had with the expression \((2)\) that:
$$ x = \frac{yr}{h} \qquad (2) $$
Consequently, we inject \((2)\) into \((4)\) and:
$$ dl = d\left(\sqrt{\left (\frac{yr}{h}\right)^2 + \ y^2}\right) $$
$$ dl = d\left(\sqrt{\frac{y^2 r^2}{h^2} + \ y^2}\right) $$
$$ dl = d\left(y^2 \left ( \sqrt{\frac{r^2}{h^2} + 1 }\right) \right) $$
$$ dl = \left ( \sqrt{\frac{r^2}{h^2} + 1 }\right) dy \qquad (5) $$
At this stage, by injecting \((5)\) into \((3)\), we obtain an integrand which is only a function of \(y\):
$$ S_{axis}(y) = \int_0^{h} dS = \frac{2 \pi r}{h} \int_0^{h} y \left ( \sqrt{\frac{r^2}{h^2} + 1 }\right) dy $$
We take out some constants again:
$$ S_{axis}(y) = \frac{2 \pi r}{h} \left ( \sqrt{\frac{r^2}{h^2} + 1 }\right) \int_0^{h} y \ dy $$
And we integrate it:
$$ S_{axis}(y) = \frac{2 \pi r}{h} \left ( \sqrt{\frac{r^2}{h^2} + 1 }\right) \Biggl[ \frac{y^2}{2} \Biggr]_0^h $$
After integration, all that remains are constants.
$$ S_{axis} = \frac{2 \pi r}{h} \left ( \sqrt{\frac{r^2}{h^2} + 1 }\right) \Biggl( \frac{h^2}{2} - \frac{0^2}{2} \Biggr) $$
$$ S_{axis} = \frac{2 \pi r}{h} \left ( \sqrt{\frac{r^2}{h^2} + 1 }\right) \frac{h^2}{2} $$
$$ S_{axis} = \pi r h \left ( \sqrt{\frac{r^2}{h^2} + 1 }\right) $$
$$ S_{axis} = \pi r h \left ( \sqrt{\frac{r^2+ h^2}{h^2} }\right) $$
$$ S_{axis} = \pi r \sqrt{ r^2+ h^2} $$
Finally, we have as a total surface,
$$ S_{cone} = S_{base} + S_{axis} $$
$$ S_{cone} = \pi r ^2 + \pi r \sqrt{ r^2+ h^2} $$
Let \( r\) be the radius of a circle represented in a coordinate system \((O, \overrightarrow{x}, \overrightarrow{y})\), and \( R\) the radius of a sphere represented in a coordinate system \((O, \overrightarrow{x}, \overrightarrow{y}, \overrightarrow{z})\). Both radii \( r\) and \( R\) have the same lengths.
As well, let us consider an infinitesimal vertical length \( dz \), such as the following figure:
The two radii \(r\) and \(R\) are respectively mobile according to \(x, y\) and \(x, y, z\).
With the Pythagorean theorem, we do have:
$$ \Biggl \{ \begin{align*} r^2 = x^2 + y^2 \\ R^2 = x^2 + y^2 + z^2 \end{align*} $$
So,
$$ r = \sqrt{R^2 - z^2} \qquad (1) $$
Considering now a moving surface of the circle as a function of \(r\), we do have:
$$ S(r) = \pi r^2 \qquad (2) $$
Injecting \((1)\) into \((2)\):
$$ S(z) = \pi (R^2 - z^2) \qquad (3)$$
We then have an infinitesimal volume \( dV\), shown in blue in the figure below.
$$ dV = S(z).dz \qquad (4) $$
Now, injecting \((3)\) into \((4)\):
$$ dV =\pi (R^2 - z^2).dy $$
We will integrate the axis \(y \) on \( [0, R] \), which will cover the volume of half a sphere.
$$ \int_0^{R} dV = \pi \int_0^{R} (R^2 - z^2) .dz $$
$$V_{\frac{1}{2}} = \pi \Bigl[ R^2z - \frac{z^3 }{3}\Bigr]_0^{R} $$
$$V_{\frac{1}{2}} = \pi \Bigl( R^3 - \frac{R^3 }{3} - (0 - 0 ) \Bigr) $$
$$V_{\frac{1}{2}} = \pi \Biggl( \frac{2R^3 }{3} \Biggr) $$
To have the complete volume, we will simply double this value.
$$ V_{sphere} = 2V_{\frac{1}{2}} $$
And as a result,
$$ V_{sphere} = \frac{4}{3}\pi R^3 $$
Let us consider a cone of height \(h\) with a circular base of radius \( r\).
For simplicity of demonstration, we have turned this cone upside down.
Likewise, we added a moving height \(y\), as well as a mobile radius \(x\).
The moving radius \(x\) then varies depending on \(y\).
By applying the Thales' theorem, we do have:
$$ \frac{x}{r} = \frac{y}{h} \Longleftrightarrow x = \frac{yr}{h} \qquad (1) $$
Now, we know that the surface of a circle of radius \(R\) is worth:
$$ S_{cercle} = \pi R^2 $$
So in the case of our mobile radius \(x\), let us call the mobile surface \(S(x)\) :
$$ S(x) = \pi x^2 \qquad (2) $$
By injecting \((2)\) into \((1)\), we obtain the surface of this same circle, this time a function of \(y\):
$$ S(y) = \pi \left( \frac{yr}{h} \right)^2 $$
$$ S(y) = \pi \frac{y^2r^2}{h^2} \qquad (3) $$
The total volume will then be the addition of all these mobile surfaces, along the vertical axis up to \(h\).
Let us call \(dV\) an element of infinitesimal volume, we do have:
$$ dV= S(y) \ dy $$
Now, injecting the value of \(S(y)\)thanks to\((3)\):
$$ \int_0^{h} dV = \int_0^{h} \frac{\pi y^2r^2}{h^2} \ dy $$
We take out some constants:
$$ V_{cone} = \int_0^{h} dV = \frac{\pi r^2}{h^2} \int_0^{h} y^2 \ dy $$
We do then have:
$$ V_{cone}(y) = \frac{\pi r^2}{h^2} \int_0^{h} y^2 \ dy $$
And now we can integrate it:
$$ V_{cone} = \frac{\pi r^2}{h^2} \Biggl( \frac{h^3}{3} -\frac{0^3}{3} \Biggr) $$
$$ V_{cone} = \frac{\pi r^2h}{3} $$
And as a result,
$$ V_{cone} = \frac{\pi r^2 h}{3} $$
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