So, I am calculating the volume by revolving the region between the curves $$2x+3y=6\quad\text{and}\quad (y-1)^2=4-x$$
about
$i)$ $x=-5$
$ii)$ $y=0$
The problem is though I can't find the intersection points. I get an unbounded integral if I try to do that and I don't think that's that right approach...
Can some give me a suggestion as to how I would go about doing this problem? Thanks!
The region is given below.
The intersection occur at the points where $y=0$ and $y=\frac{7}{2}$. Let us write $$g(y)=4-(y-1)^2=-y^2+2y+3$$ and $$f(y)=\frac{6-3y}{2}.$$ If the region $R$ is revolved about the line $x=-5$, then the volume of the solid generated (by using the washer method) is given by $$V=\int_{0}^{\frac{7}{2}}\pi\bigg(\big[g(y)-(-5)\big]^2-\big[f(y)-(-5)\big]^2\bigg)dy$$ and if the region $R$ is revolved about the line $y=0$ then the volume of the solid generated (by using the shell method) is given by $$V=\int_{0}^{\frac{7}{2}}2\pi y\cdot\big[g(y)-f(y)\big]dy$$ Can you proceed with the computations?