Question

The expression $V={\int }_0^H\pi R^2(1-h/H{)}^{2}dh$ for the volume of a cone is equal to

• A. ${\int }_0^R\pi R^2(1-h/H{)}^{2}dr$
• B. ${\int }_0^R\pi RH{(1-\frac{r}{R})}^{2}dr$
• C. ${\int }_0^{H}2\pi rH(1-r/R)dh$
• D. ${\int }_0^{H}2\pi rH{(1+\frac{r}{R})}^{2}dr$

Answer 1

The correct option is B ${\int }_0^R\pi RH{(1-\frac{r}{R})}^{2}dr$

Method I:
Given $V={\int }_0^H\pi R^2(1-h/H{)}^{2}dh$
Put $1-\frac{h}{H}=t$
$\Rightarrow =-Hdt$ and $h=0,t=1$, while at $h=H,t=0$
So, ${\int }_0^R\pi RH{(1-\frac{r}{R})}^{2}dr$
$={\int }_1^0\pi R^2{t}^2(-Hdt)=-\pi R^{2}H{\left(\frac{t^3}{3}\right)}_1^0$
$\frac{1}{3}\pi R^{2H}$ i.e., volume of cone
Similarly let us take option (b)
${\int }_0^R\pi RH{(1-\frac{r}{R})}^{2}dr$
Put $1-\frac{r}{R}=t\Rightarrow dr=-Rdt$
So $v={\int }_0^R\pi RH{(1-\frac{r}{R})}^{2}dr$
$={\int }_0^R\pi RH(-Rdt)=\frac{1}{3}\pi R^{2}H$
i.e., volume of cone
Hence option (b) is correct
Method II:
${\int }_0^H\pi R^2{(1+\frac{r}{H})}^{2}dh$
From figure,
$\frac{r}{R}=\frac{h}{H}\Rightarrow h=\frac{H}{R}.r$


$Rightarrowdh=\frac{H}{R}dr$
$V={\int }_0^R\pi R^2{(1-\frac{r}{R})}^2.\frac{H}{R}dr$
$={\int }_0^R\pi RH{(1-\frac{r}{R})}^{2}dr$