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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

$$\frac{h_1}{h_1+h}=\frac{r}{R}\Rightarrow h_1=\frac{rh}{R-r}$$$$bigcone:$$$$V_{big}=\frac{\pi R^2(h+h_1)}{3}=\frac{\pi R^{3}h}{3(R-r)}$$$$smallcone:$$$$V_{small}=\frac{\pi r^2{h}_1}{3}=\frac{\pi r^{3}h}{3(R-r)}$$$$frustum:$$$$V=V_{big}-V_{small}=\frac{\pi R^{3}h}{3(R-r)}-\frac{\pi r^{3}h}{3(R-r)}$$$$=\frac{\pi h(R^3-r^3)}{3(R-r)}=\frac{\pi h(R^2+Rr+r^2)}{3}$$

Commented by Mingma last updated on 23/Feb/23

Perfect!

Commented by otchereabdullai last updated on 24/Feb/23

$$supperprofW$$

Commented by Humble last updated on 01/Mar/23

$$superb!$$

Commented by Humble last updated on 01/Mar/23

$$superb!$$

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

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

$$y=r+\frac{(R-r)x}{h}$$$$dV=\pi y^{2}dx$$$$V={\int }_0^h\pi y^{2}dx=\pi {\int }_0^h[r^2+\frac{2r(R-r)x}{h}+\frac{{(R-r)}^2{x}^2}{h^2}]dx$$$$=\pi {[r^{2}x+\frac{r(R-r)x^2}{h}+\frac{{(R-r)}^2{x}^3}{3h^2}]}_0^h$$$$=\pi [r^{2}h+\frac{r(R-r)h^2}{h}+\frac{{(R-r)}^2{h}^3}{3h^2}]$$$$=\pi h[r^2+r(R-r)+\frac{{(R-r)}^2}{3}]$$$$=\pi h\left[\frac{3r^2+3rR-3r^2+R^2-2Rr+r^2}{3}\right]$$$$=\frac{\pi h(R^2+Rr+r^2)}{3}✓$$

Commented by Mingma last updated on 23/Feb/23

Nice solution, sir!

Commented by Mingma last updated on 23/Feb/23

Nice sir!

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