Lynn’s answer was correct
F(x) = -2x^2 +32
F(3) =-2(3)^2 +32
F(3)=-2(9)+32
F(3) =-18+32
F(3) = 14
X=14
The answer is C.
|-4+4| < 1
|0| < 1
0 < 1
Solution :
Along the edge 
The parametric equation for
is given :

Along edge 
The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain
is then given by :


Along edge 
The parametric equation for
is :

Now,
x = 9t, ⇒ dx = 9 dt
y = 0, ⇒ dy = 0

And


Then :

![$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$](https://tex.z-dn.net/?f=%24%3D%5Cint_0%5E1%20%5Cleft%5B%5Cleft%28%209%20%5Csin%20%5Cfrac%7B%5Cpi%7D%7B2%7Dt%5Cright%29%5E2%5Cleft%289%20%5Ccos%20%5Cfrac%7B%5Cpi%7D%7B2%7Dt%5Cright%29%5Cleft%28-%5Cfrac%7B7%20%5Cpi%7D%7B2%7D%20%5Csin%20%5Cfrac%7B%5Cpi%7D%7B2%7Dt%20dt%5Cright%29%20%2B%20%5Cleft%28%209%20%5Ccos%20%5Cfrac%7B%5Cpi%7D%7B2%7Dt%5Cright%29%5E2%5Cleft%289%20%5Csin%20%5Cfrac%7B%5Cpi%7D%7B2%7Dt%5Cright%29%5Cleft%28%5Cfrac%7B7%20%5Cpi%7D%7B2%7D%20%5Ccos%20%5Cfrac%7B%5Cpi%7D%7B2%7Dt%20dt%5Cright%29%20%5Cright%5D%24)
![$=\left[-9^4\ \frac{\cos^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} -9^4\ \frac{\sin^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} \right]_0^1$](https://tex.z-dn.net/?f=%24%3D%5Cleft%5B-9%5E4%5C%20%5Cfrac%7B%5Ccos%5E4%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7Dt%5Cright%29%7D%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20-9%5E4%5C%20%5Cfrac%7B%5Csin%5E4%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7Dt%5Cright%29%7D%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%20%5Cright%5D_0%5E1%24)
= 0
And
x = 0, ⇒ dx = 0
y = 9 t, ⇒ dy = 9 dt

Therefore,

= 0 + 0 + 0
Applying the Green's theorem


Here,



Therefore,


The vector field F is =
is conservative.
The region in question is the set

or equivalently,
![R = \left\{ (x, y) : 0 \le y \le 1 \text{ and } 0 \le x \le \sqrt[3]{y} \right\}](https://tex.z-dn.net/?f=R%20%3D%20%5Cleft%5C%7B%20%28x%2C%20y%29%20%3A%200%20%5Cle%20y%20%5Cle%201%20%5Ctext%7B%20and%20%7D%200%20%5Cle%20x%20%5Cle%20%5Csqrt%5B3%5D%7By%7D%20%5Cright%5C%7D)
Cross sections are taken perpendicular to the y-axis, which means each section has a base length equal to the horizontal distance between the curve y = x³ and the line x = 0 (the y-axis). This horizontal distance is given by
y = x³ ⇒ x = ∛y
so that each triangular cross section has side length ∛y.
The area of an equilateral triangle with side length s is √3/4 s², so each cross section contributes an infinitesimal area of √3/4 ∛(y²).
Then the volume of this solid is
![\displaystyle \frac{\sqrt3}4 \int_0^1 \sqrt[3]{y^2} \, dy = \frac{\sqrt3}4 \int_0^1 y^{2/3} \, dy = \frac{\sqrt3}4\cdot\frac35 y^{5/3} \bigg|_0^1 = \boxed{\frac{3\sqrt3}{20}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Csqrt3%7D4%20%5Cint_0%5E1%20%5Csqrt%5B3%5D%7By%5E2%7D%20%5C%2C%20dy%20%3D%20%5Cfrac%7B%5Csqrt3%7D4%20%5Cint_0%5E1%20y%5E%7B2%2F3%7D%20%5C%2C%20dy%20%3D%20%5Cfrac%7B%5Csqrt3%7D4%5Ccdot%5Cfrac35%20y%5E%7B5%2F3%7D%20%5Cbigg%7C_0%5E1%20%3D%20%5Cboxed%7B%5Cfrac%7B3%5Csqrt3%7D%7B20%7D%7D)
I've attached some sketches of the solid with 16 and 64 such cross sections to give an idea of what this solid looks like.
Answer:
12 20 22
Step-by-step explanation: