(intergation) area of region bounded by a curve and a line with unknown a Let $S1$ be the area of the region bounded by the curve $y = x^2$ and the lines $y = ax, (0 < a < 1)$, and let $S2$ be the area of the region bounded by the curve $y = x^2$, the lines $y = ax, (0 < a < 1)$ and the line $x = 1$, where $a$ is the same constant.
(a) Find the value of $a$ that minimizes the value of $S1 + S2$. What is the minimum value?
(b) Find the volume of the solid generated by revolving the region, that takes the minimum value of $S1 + S2$, about the x-axis.
I find $S1 = \displaystyle \frac{a^3}{6}$ and $S2 = \displaystyle \frac{a^3}{3} - \frac{a}{2} + \frac{1}{3}$, but how can I calculate the value of $a$?

You have calculated value of $\small S_1$ and $\small S_2$ correctly. Now to find value of $a$ that minimizes $ \small S = S_1+S_2$, take the derivative of $S$ with respect to $a$ and equate to zero.
$\small S = S_1 + S_2 = \displaystyle \frac{a^3}{3} - \frac{a}{2} + \frac{1}{3}$
$\small S' = \displaystyle a^2 - \frac{1}{2} = 0 \implies a = \frac{1}{\sqrt2}$
That leads to minimum area of $\small S_1 + S_2 = \displaystyle \frac{\sqrt2 - 1}{3 \sqrt2}$
Now you need to find volume of solid generated by revolution of the region about x-axis $\big($for $a = \displaystyle \small \frac{1}{\sqrt2}\big)$. For $\small S_1$, both shell and washer methods are fine. For $\small S_2$, it is easier to apply washer method as the shell method will require you to split the integral into two.
