Question-187921

Question Number 187921 by Mingma last updated on 23/Feb/23

Answered by HeferH last updated on 23/Feb/23

V_{1} - V_{2} = V_{3}

Answered by mr W last updated on 23/Feb/23

Commented by mr W last updated on 23/Feb/23

\frac{h_{1}}{h_{1} + h} = \frac{r}{R} \Rightarrow h_{1} = \frac{r h}{R - r}

b i g c o n e :

V_{b i g} = \frac{π R^{2} (h + h_{1})}{3} = \frac{π R^{3} h}{3 (R - r)}

s m a l l c o n e :

V_{s m a l l} = \frac{π r^{2} h_{1}}{3} = \frac{π r^{3} h}{3 (R - r)}

f r u s t u m :

V = V_{b i g} - V_{s m a l l} = \frac{π R^{3} h}{3 (R - r)} - \frac{π r^{3} h}{3 (R - r)}

= \frac{π h (R^{3} - r^{3})}{3 (R - r)} = \frac{π h (R^{2} + R r + r^{2})}{3}

Commented by Mingma last updated on 23/Feb/23

Perfect!

Commented by otchereabdullai last updated on 24/Feb/23

s u p p e r p r o f W

Commented by Humble last updated on 01/Mar/23

s u p e r b!

Commented by Humble last updated on 01/Mar/23

s u p e r b!

Answered by mr W last updated on 23/Feb/23

Commented by mr W last updated on 23/Feb/23

y = r + \frac{(R - r) x}{h}

d V = π y^{2} d x

V = \int_{0}^{h} π y^{2} d x = π \int_{0}^{h} [r^{2} + \frac{2 r (R - r) x}{h} + \frac{{(R - r)}^{2} x^{2}}{h^{2}}] d x

= π {[r^{2} x + \frac{r (R - r) x^{2}}{h} + \frac{{(R - r)}^{2} x^{3}}{3 h^{2}}]}_{0}^{h}

= π [r^{2} h + \frac{r (R - r) h^{2}}{h} + \frac{{(R - r)}^{2} h^{3}}{3 h^{2}}]

= π h [r^{2} + r (R - r) + \frac{{(R - r)}^{2}}{3}]

= π h [\frac{3 r^{2} + 3 r R - 3 r^{2} + R^{2} - 2 R r + r^{2}}{3}]

= \frac{π h (R^{2} + R r + r^{2})}{3} ✓

Commented by Mingma last updated on 23/Feb/23

Nice solution, sir!

Commented by Mingma last updated on 23/Feb/23

Nice sir!