Volume of a right circular cone

[MathsLearner]

I tried to interpret in the above way, the equation of the line is
[math] y = \frac r h x; \\ \int_0^r (2\pi y) \frac h r ydy \\ 2\pi \frac h r \int_0^r y^2 dy \\ \frac {2\pi r^2 h} 3 [/math]I am getting an extra of 2. What is the mistake?

 

[lev888]

Where does 2 come from in the first place? What formula for volume are you using? Which axis is it revolving about?

 

[MarkFL]

I agree with the equation of the line. If you're going to use the shell method, then:

[MATH]dV=2\pi y(h-x)\,dy=2\pi y\left(h-\frac{h}{r}y\right)\,dy=2\pi \frac{h}{r} y(r-y)\,dy[/MATH]
Hence:

[MATH]V=2\pi \frac{h}{r}\int_0^r y(r-y)\,dy=2\pi \frac{h}{r}\int_0^r ry-y^2\,dy=2\pi \frac{h}{r}\left[r\frac{y^2}{2}-\frac{y^3}{3}\right]_0^r=2\pi \frac{h}{r}\cdot\frac{r^3}{6}=\frac{\pi}{3}hr^2[/MATH]
I'd actually use the disk method though, where:

[MATH]dV=\pi y^2\,dx=\pi\left(\frac{r}{h}x\right)^2\,dx=\pi\frac{r^2}{h^2}x^2\,dx[/MATH]
Hence:

[MATH]V=\pi\frac{r^2}{h^2}\int_0^h x^2\,dx=\frac{\pi r^2}{3h^2}\left[x^3\right]_0^h=\frac{\pi r^2}{3h^2}\cdot h^3=\frac{\pi}{3}hr^2[/MATH]

 

[MathsLearner]

Where does 2 come from in the first place? What formula for volume are you using? Which axis is it revolving about?

It is revolving about the X-axis

As MarkFl pointed out i understood the mistake
The radius [math] 2\pi ydy [/math] the height is [math] h - \frac {hy} r[/math][math] \int_0^r 2\pi y (h - \frac {hy} r) dy = \pi \frac {hr^2} 3 [/math]

 

[Steven G]

Watch this video from nothing but math proofs on youtube