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Using integration find the volume of cutted cone

Amer

Active member
Mar 1, 2012
275
if we cut a right cone parallel to the base having a two radius r and R The picture
Mas.jpg

I want to use the volume of revolution around the y-axis
we have the line

[tex]y - 0 = \dfrac{h}{r-R} (x - R)[/tex]
[tex] x = \frac{r-R}{h} y +R [/tex]

The volume will be
[tex] \pi \int_{0}^{h} \left(\frac{r-R}{h} y + R\right)^2 dy [/tex]
[tex]\pi \int_{0}^{h} \frac{(r-R)^2y^2}{h^2} + \frac{2R(r-R)y}{h} + R^2 dy [/tex]
now i just have to evaluate the integral, did i miss something ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Looks correct to me, however, I would use a substitution to simplify matters:

$\displaystyle V=\pi\int_0^h\left(\frac{r-R}{h}y+R \right)^2\,dy$

Let:

$\displaystyle u=\frac{r-R}{h}y+R\,\therefore\,du=\frac{r-R}{h}\,dy$

hence:

$\displaystyle V=\frac{h}{r-R}\pi\int_{R}^{r}u^2\,du$