Question

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

check the picture below.

thus, if one side on the square is x - x², then the area for the square is (x - x²)(x - x²), or (x - x²)²

now... what is the "x" value for the region, let's check their intersection.

$\bf \begin{cases} y=x\\ y=x^2 \end{cases}\implies x=x^2\implies 0=x^2-x\implies 0=x(x-1) \\\\\\ x= \begin{cases} 0\\ 1 \end{cases}\\\\ -------------------------------\\\\ \displaystyle \int\limits_{0}^{1}\ (x-x^2)^2\cdot dx\implies \int\limits_{0}^{1}\ (x^2-2x^3+x^4)\cdot dx \\\\\\ \cfrac{x^2}{2}-\cfrac{2x^4}{4}+\cfrac{x^5}{5}\implies \left. \cfrac{x^2}{2}-\cfrac{x^4}{2}+\cfrac{x^5}{5} \right]_{0}^{1}\implies \left[ \cfrac{1}{2}-\cfrac{1}{2}+\cfrac{1}{5} \right]-[0]\implies \cfrac{1}{5}$

Answer 2

Answer:  Volume (V) = 1/30

Step-by-step explanation:

Pictorial image can be seen below.

The Volume of the solid is evaluated as,

Volume (V) = ∫_(x=0)^1▒〖(x-x∧2)∧2dx〗

V = ∫_(x=0)^1▒〖(x∧2+x∧4-2x∧3)dx〗

V = [(x∧3)/3+(x∧5)/5-2((x∧4)/4) ]x=0 to 1  

V = [1/3 + 1/5 -1/2]

V = (10+6-15)/30  

V = 1/30

Therefore, the volume of the solid is  

Volume (V) = 1/30

I hope this helps, Cheers.