I'm asked to find the volume of a container of height $13.5$ The container is made by rotating $y = 1.5x^2$ where $0\leq x \leq 3$ about the axis $x=-0.5$ (the bottom is flat).
I tried finding the right volume a number of ways, but I believe my most solid attempt so far is this:
I invert the function to get $y = \sqrt{\frac{2}{3}x}$.
From this I tried to calculate $\pi * \int_{0}^{3} \sqrt{\frac{2}{3}x} dx$ which gave me $3\pi$ which I thought was correct, but it wasnt. Plotting this, I realized the height becomes 3 when I integrate the inverted function from 0 to 3, so I thought I should maybe integrate from 0 to 13.5. In that case I get $\frac{243\pi}{4}$ which is also incorrect. I'm sure I've misinterpreted what it means to rotate the function about the axis $x=-0.5$... Any help is greatly appreaciated!
I translated $0.5$ units to the right so I had the equation
$$y=\frac{3}{2} \left(x-\frac{1}{2}\right)^2$$ to rotate around $y$ axis and I got $$V=\pi \int_0^3 \left(\frac{3}{2} \left(x-\frac{1}{2}\right)^2\right)^2 \, dx=\frac{14067 \pi }{320}\approx 138.1$$ Hope this is useful