A simple guy: AMAZING DAY

Today i go to internet and search some information for mathematics.After learning the integration in additional mathematics,i feel very excited and enjoyed because i had learnt something new.INTEGRATION is important in the field of calculus,and more broadly,mathematical analysis.Integration is quite a tough chapter but if you know the concepts ,it will be easy....After coming back from Jen Weng's house,i got nothing to do.Hence,i go to internet and search something that are relevant to science and mathematics.It is fun......but i realize somethings...actually our integration in SPM is not even a basic or foundation..the internet info about integration are damn crazy and hard.....STUPID....and i also discover that integration concepts are also involve in volume formula deviation..Here are some example......

VOLUME OF SPHERE

The volume of a sphere is the integral of infinitesimal circular slabs of thickness $dx$ . The calculation for the volume of a sphere with center 0 and radius $r$ is as follows.

The radius of the circular slabs is $y = \sqrt{r^2-x^2}$

The surface area of the circular slab is $πy 2$ .

The volume of the sphere can be calculated as $\int_{-r}^r \pi(r^2-x^2) \,dx = \int_{-r}^r \pi r^2\,dx - \int_{-r}^r \pi x^2 \,dx$

Now $\int_{-r}^r \pi r^2\,dx = 2\pi r^3$

and $\int_{-r}^r \pi x^2 \,dx = 2 \pi \frac{r^3}{3}$

Combining yields $\left(2-\frac{2}{3}\right)\pi r^3 = \frac{4}{3}\pi r^3$

This formula can be derived more quickly using the formula for the sphere's surface area, which is $4πr 2$ . The volume of the sphere consists of layers of infinitesimal spherical slabs, and the sphere volume is equal to

$\int_0^r 4\pi r^2 \,dr$ = $\frac{4}{3}\pi r^3$

VOLUME OF CONE

The volume of a cone is the integral of infinitesimal circular slabs of thickness $d x$ . The calculation for the volume of a cone of height $h$ , whose base is centered at (0,0) with radius $r$ is as follows.

The radius of each circular slab is $\begin{cases} r, & \mbox{if }\mbox{ x=0} \\ 0, & \mbox{if }\mbox{ x=h} \end{cases}$ , and varying linearly in between -- that is, $r\frac{(h-x)}{h}$

The surface area of the circular slab is then $\pi \left(r\frac{(h-x)}{h}\right)^2 = \pi r^2\frac{(h-x)^2}{h^2}$

The volume of the cone can then be calculated as $\int_{0}^h \pi r^2\frac{(h-x)^2}{h^2} dx$

Substituting $u = h - x$ gives an integral with reversed limits, and $d x = - d u$ : that is, $\int_{h}^0 \pi r^2\frac{u^2}{h^2} (-du)$

Swapping the limits makes this $-\int_{0}^h \pi r^2\frac{u^2}{h^2} (-du) = \int_{0}^h \pi r^2\frac{u^2}{h^2} du = \frac{\pi r^2}{h^2}\int_{0}^h u^2 du$