Volume of Revolution: Volume of resulting solid

Question

The triangular region bounded by the lines $x+y=1$, $x=0$, and $y=0$ is rotated around the line $x=-3$. What is the volume of the resulting solid?

I have found similar questions asked but they all seem to be rotated horizontally. I did the problem and got $\;28{\pi\over 3}$, but I think my limits of integration might be wrong.

Answer 1

if you consider a thin disc of thickness $dy$ & radius $r=x(y)+3$ about the line $x=-3$ & subtract the volume of corresponding thin concentric disc $\pi (3)^2dy$ then the required volume of solid of rotation
$$=\int_0^1(\pi(x+3)^2-\pi(3^2))\ dy$$ $$=\pi\int_0^1((-y+1+3)^2-9)\ dy$$ $$=\pi\int_0^1(y^2-8y+7)\ dy$$ $$=\frac{10\pi}{3}$$ hope answer is correct

Answer 2

Let $X=x+3$, then the volume of the triangular region bounded by the lines $x+y=1$, $x=0$, and $y=0$ is rotated around the line $x=-3$ is equal to the volume of the triangular region bounded by the lines $X=4-y$, $X=3$, and $y=0$ is rotated around the line $X=0$.

Are you able to apply the usual formula now? $$\mbox{Volume}=\pi\int_{y=a}^b(f^2(y)-g^2(y)) dy.$$