1. Bounded Solid

A base of a solid is bounded by x+2y=4 and coordinate axes. What is volume of solid if every cross section perpendicular to x-axis is a semicircle?

a) 2pi/3 b) 4pi?3 c) 8pi/3 d) 32pi/3 e) 64pi/3

Please help me with this calculus problem. I am so lost in my class. step by step work would be greatly appreciated. The work does not have to be completely accurate, I just need the steps. Thank you in advance.

2. Imagine slicing the solid into many very thin slices, perpendicular to the x-axis. If the slices are thin enough that the width of the slices does not change much, it's volume is, approximately, the area of the base times the thickness. You are told that "every cross section perpendicular to x-axis is a semicircle" so the area of each slice is $\displaystyle \frac{1}{2}\pi r^2$. The line segment form y= 0 to x+ 2y= 4, has length y= (4- x)/2 and forms a diameter of the semicircle. That is, the area of each slice is $\displaystyle \frac{1}{8}\pi (4- x)^2$. Since the thickness is along the x-axis, we call the thickness "$\displaystyle \Delta x$". The volume of each slice, area times thickness, is $\displaystyle \frac{1}{8}\pi (4- x)^2 \Delta x$. We can approximate the volume of the solid by adding the volumes of all slices: $\displaystyle \sum\frac{1}{8}\pi (4- x)^2\Delta x= \frac{\pi}{8} \sum (4- x)^2$. That's a "Riemann sum" and we can get the exact volume by taking the limit with more and more slices of smaller and smaller thickness, giving the integral
$\displaystyle \frac{\pi}{8}\int_0^4(4- x)^2 dx= \frac{\pi}{8}\int_0^4 (16- 8x+ x^2)dx$.

That lower bound, 0, is where the line crosses the y-axis (so x= 0) and the upper bound, 4, is where the line, x+ 2y= 4, crosses the x-axis (so y= 0).