6. set up an integral for the volume of the solid s generated by rotating the region r bounded by $x = 4y$…

Explanation:

Step1: Find intersection points

Set $x=4y$ and $y=x^\frac{1}{3}$ (so $x=y^3$) equal: $$4y = y^3$$ $$y^3 - 4y = 0$$ $$y(y^2 - 4) = 0$$ $$y(y-2)(y+2)=0$$ Solutions: $y=-2, 0, 2$. Corresponding $x$-values: $x=-8, 0, 8$. The bounded region uses $y \in [0,2]$ (since $y=-2$ gives a separate intersection).

Step2: Express radii for washer method

Rotating about $y=2$. For a given $y$, outer radius $R(y) = 2 - y$ (distance from $y$ to $y=2$), inner radius $r(y) = 2 - \frac{x}{4} = 2 - \frac{y^3}{4}$ (since $x=y^3$ so $x/4 = y^3/4$).

Step3: Set up volume integral

Washer method volume formula: $V=\pi\int_{a}^{b}\left([R(y)]^2 - [r(y)]^2\right)dy$ $$V=\pi\int_{0}^{2}\left[(2 - y)^2 - \left(2 - \frac{y^3}{4}\right)^2\right]dy$$

(Sketch description: Plot $x=4y$ (a straight line through origin with slope 1/4) and $y=x^\frac{1}{3}$ (a curve passing through origin, increasing concave down). The region R is enclosed between these two curves from $x=0$ to $x=8$, or $y=0$ to $y=2$.)

Answer:

$$V=\pi\int_{0}^{2}\left[(2 - y)^2 - \left(2 - \frac{y^3}{4}\right)^2\right]dy$$ (Region sketch: Enclosed area between $x=4y$ and $y=x^\frac{1}{3}$ from $(0,0)$ to $(8,2)$)