Find the volume of the solid of revolution obtained by revolving the first quadrant plane bounded by $y= -\frac{1}{4}x^2 + x, y= -\frac{1}{8}x^2 + x,$ and the $x-$axis about the $y-$axis.
Attempt: First of all, $$y= -\frac{1}{4}x^2 + x \implies -4y = x^2-4x \iff 4-4y = x^2-4x+4=(x-2)^2 \iff x=2+\sqrt{4-4y}$$ and $$y= -\frac{1}{8}x^2 + x \implies -8y = x^2-8x \iff 16-8y = x^2-8x+16=(x-4)^2 \iff x=4+\sqrt{16-8y}.$$ By considering the graphic, we have $$V=\pi \int_0^2 \left((4+\sqrt{16-8y})^2 - (2+\sqrt{4-4y})^2 \right) \ dy.$$ But, this will give us an imaginary number.
How to handle this one? Thanks in advanced.
There is no reason to use the washer method. Instead, the method of cylindrical shells is appropriate. We would simply let a representative shell have radius $x$, height $(-\frac{1}{8} x^2 + x) - \max(-\frac{1}{4} x^2 + x, 0)$, and differential thickness $dx$. When $x \in [0,4]$, the function $-\frac{1}{4}x^2 + x$ is nonnegative; when $x \in (4,8]$, it is negative so we use the lower bound of the $x$-axis as indicated in the problem. So we would split the interval of integration into these two subintervals and write
$$V = \int_{x=0}^4 2\pi x \left(-\frac{1}{8} x^2 + x + \frac{1}{4}x^2 - x \right) \, dx + \int_{x=4}^8 2\pi x \left(-\frac{1}{8} x^2 + x \right) \, dx. \tag{1}$$
The details of the computation of $(1)$ is left to the reader.
If one insists on using washers, how would we set up the volume integral correctly? This requires an understanding of the multivalued nature of the inverse function. When $y \in (1,2]$, the washer is bounded only by the curve $y = -\frac{1}{8} x^2 + x$ because the other curve does not reach $y = 1$. So this part of the volume is given by $$V_1 = \int_{y=1}^2 \pi \left( (4 + 2 \sqrt{4-2y})^2 - (4 - 2 \sqrt{4-2y})^2 \right) \, dy, \tag{2}$$ since the outer radius is the larger root of the quadratic $-\frac{1}{8} x^2 + x - y = 0$, and the inner radius is the smaller root of the same quadratic.
However, when $y \in [0,1]$, the washer also intersects the lower parabolic curve $y = -\frac{1}{4} x^2 + x$. This will cause there to be two distinct concentric washers, not just one, because each parabolic curve creates two boundaries. Computing the inverse of both curves and arranging them in ascending value, we have $$4 - 2 \sqrt{4-2y} \le 2 - 2 \sqrt{1-y} \le 2 + 2 \sqrt{1-y} \le 4 + 2 \sqrt{4-2y}. \tag{3}$$ So this part of the volume integral would be written as $$V_2 = \int_{y=0}^1 \pi \left((4 + 2 \sqrt{4-2y})^2 - (2 + 2 \sqrt{1-y})^2 + (2 - 2 \sqrt{1-y})^2 - (4 - 2 \sqrt{4-2y})^2\right) \, dy. \tag{4}$$ Therefore $V = V_1 + V_2$ using the washer method, and again I leave the details of the computation to the reader.