well, first off, when it comes to volumes by rotation, we'd want to graph them, Check the picture below. Our rotation over the axis will give us a "washer", so we'll be using the washer method.
now, our axis of rotation is the y-axis or namely x = 0, x = 0 is a vertical line, meaning we have to put the functions in y-terms, that is in f(y) form
[tex]y = x^2\implies \pm\sqrt{y}=x\implies \boxed{\pm\sqrt{y}=f(y)} \\\\\\ y = 4x\implies \cfrac{y}{4}=x\implies \boxed{\cfrac{y}{4}=g(y)}[/tex]
if we look at the picture, the parabola is the farthest from the axis of rotation and the line is the closest, or namely R² and r² respectively.
the way I get the area for R² and r², is by using the same I do with "area under the curve", so if I say call the axis of rotation h(y), the way I get the area is
R => f(y) - h(y)
r => g(y) - h(y)
so let's proceed.
[tex]\textit{area under the curve}\\ \begin{array}{llll} \sqrt{y}- 0\implies &\sqrt{y}\\ \frac{y}{4}-0\implies &\frac{y}{4} \end{array}\qquad \qquad \begin{array}{llll} \stackrel{R^2}{(\sqrt{y})^2}-\stackrel{r^2}{\left( \frac{y}{4} \right)^2}\\[-0.5em] \hrulefill\\ y\qquad -\qquad \frac{y^2}{16} \end{array}[/tex]
now, we want to get the area enclosed by both, and thus we'd need their points of intersection, setting both to [tex]\sqrt{y}=\cfrac{y}{4}[/tex] which in short gives us the bounds of 0 and 16.
[tex]\pi \displaystyle\int_{0}^{16}\left( y - \cfrac{y^2}{16} \right)dy\implies \pi \int_{0}^{16}y\cdot dy-\pi \int_{0}^{16}\cfrac{y^2}{16}\cdot dy\\\\\\\pi \cdot \left. \cfrac{y^2}{2} \right]_{0}^{16}-\pi \cdot \left. \cfrac{y^3}{48} \right]_{0}^{16}\implies 128\pi -\cfrac{256\pi }{3}\implies \boxed{\cfrac{128\pi }{3}}[/tex]
