volume of solid generated by revolving the triangle about $y$ axis

Question

Volume of solid generated by revolving the region formed by triangle $(5,0),(5,2),(7,2)$ about $y$ axis is

What i Try: Let $3$ points be $A(5,0)$ and $B(5,2)$ and $C(7,2)$

When we rotate a $\triangle ABC$ about $y$ axis. We get a cone

Whose outer radius $r_{1}=y=f(x)=x-5.$

i.e equation of line $AC$ is $y=x-5$

And here inner radius $r_{2}=5$

So volume is $$\int^{7}_{5}\pi\bigg(r^2_{1}-r^2_{2}\bigg)dx$$

$$\int^{7}_{5}\bigg((x-5)^2-5^2\bigg)dx=-\frac{142\pi}{3}$$

I did not understand where is my solution wrong and why I am getting negative answer. Please, help me.

Answer 1

Hints:

$$V_y=\pi\int\limits_0^2((y+5)^2 - 5^2)dy$$

Answer 2

What you get is a truncated cone. And the outer radius is $x$, not $x-5$. So, the volume is$$\pi\int_0^2(x+5)^2-5^2\,\mathrm dx=\frac{68\pi}3.$$

Answer 3

Hint. So if you skect the region, you see that the required figure is the difference between a cylindrical shell, and such a shell bevelled on the inside, as it were. So the integral you should consider on the interval $[5,7]$ actually is

$$\int 4πx\mathrm dx-\int 2πx(x-5)\mathrm dx.$$

The first is just a difference of cylinders, but proceeding formally too is straightforward...