Find value of area of semicircle cross-sections for the lines $x+2y=8$, the x-axis, and the y-axis The base of a solid is a region in the $1st$ quadrant bounded by the x-axis, the y-axis, and the line $x+2y=8$. If the cross-sections of the solid perpendicular to the x-axis are semicircles, what is the volume of the solid?
Area of a semicircle: $\pi r^2\over 2$
Diameter = $8-x\over 2$.
What's my error? 
 A: The general formula for finding the volume of a solid via slicing is
A = $\displaystyle \int_{a}^{b} A(x) dx$ 
where A(x) is the cross-sectional area of the solid.
We know that the cross-sections are semicircles perpendicular to the x-axis. To find the area of the typical semicircle, we must first find the typical radius.
For each semicircle, the diameter extends from the x-axis to the line x + 2y = 8. Thus to find the typical diameter we must subtract:
$\frac{8-x}{2} - 0 = \frac{8-x}{2} = diameter$. To find the radius we divide by 2: 
r = $\frac{8-x}{4}$. 
Since the area of a semicircle is $\frac{\pi r^2}{2}$, to get the area of our typical cross-section we just plug in:
A(x) = $\frac{\pi}{2}(\frac{8-x}{4})^2$
Now before we integrate we must find the limits of integration, which are in this case the intersection of the line with the x and y axes. So we have a = 0 and b = 8. 
Thus we integrate to get volume: 
V = $\displaystyle \int_{0}^{8} \frac{\pi}{2}(\frac{8-x}{4})^2 dx$ = $\frac{16\pi}{3}$
