Question

Fan and medal for calc help!!
The base of a solid in the xy-plane is the first-quadrant region bounded y = x and y = x^2. Cross sections of the solid perpendicular to the x-axis are squares. What is the volume, in cubic units, of the solid?

Answer 1

Drawing a picture shows that x is the line on top of the bounded region. Setting them equal gives x=x^2 => x=0,1 so those are the bounds. If r is the base of a semicircle, then r/2 is the radius and the full area is pi/8*r^2. Thus, the volume is integral of pi/8*(x-x^2)^2 from 0 to 1. Evaluating gives pi/48=.06545.

2. The picture looks like this http://prntscr.com/7tl3wh
Setting ln(x)=1 and ln(1)=x gives the bounds of x=1 and x=e
The top line is just y=1, which is always 2 units above from y=-1, so our first volume is pi*2^2.
The bottom line is ln(x), and it's height above y=-1 is just ln(x)+1, so our second volume is pi*(ln(x)+1)^2. The volume from our region is just the difference between these two volumes, so our integral is pi*[2^2-(ln(x)+1)^2] from x=1 to e dx.

3. I don't know the integral of arcsin(x), so I'll convert the equation to x=,,, and integrate along y. arcsinsin(4/4)=pi/2, so our bounds are 0 to pi/2. Here's the work http://prntscr.com/7tln9a

Answer 2

i'd go \(\delta V = (x - x^2)^2 ~ \delta x \)

so \(V = \int\limits_0^1 ~ (x - x^2)^2 ~ dx \)

https://www.wolframalpha.com/input/?i=int+(x+-+x%5E2)%5E2+dx+from+0+to+1

Answer 3

@slizzard Did you just try googling it..? The question doesn't pertain to semicircles.

Answer 4

Ahh so V=∫[0,1] (x−x^2)^2 dx expresses that area