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.
Best Answer
Hints:
$$V_y=\pi\int\limits_0^2((y+5)^2 - 5^2)dy$$