Question

What is the volume of the solid whose cross- sections are circular disks perpendicular to the x-axis and with diameters on the region bounded by the curves y = x and y=x^2?

Answer 1

Answer:

the volume of the solid is $\frac{\pi }{120}$

Step-by-step explanation:

Given the data in the question;

y = x and y = x²

so

x = x²

x - x² = 0

Radius will be; r = ( x - x² )/2

Area of the circular disk πr² = π[ ( x - x² )/2 ]²

A = $\frac{\pi }{4}$( x - x² )²

So, our volume will be;

V = ₀∫¹ A(x) dx

we substitute

= ₀∫¹ $\frac{\pi }{4}$( x - x² )² dx        

{ lets expand; ( x - x² )² = x(x-x²) -x²(x-x²) = x² - x³ - x³ + x⁴ = x² + x⁴ - 2x³  }

so we have;

=  ₀∫¹ $\frac{\pi }{4}$( x² + x⁴ - 2x³ ) dx  

= $\frac{\pi }{4}$ $[$ $\frac{x^3}{3}$ + $\frac{x^5}{5}$ - $\frac{2}{4} x^4$ $]^1_0$

= $\frac{\pi }{4}$[ $\frac{1}{3}$ + $\frac{1}{5}$ - $\frac{2}{4}$ ]

= $\frac{\pi }{4}$[ $\frac{1}{30}$ ]

V = $\frac{\pi }{120}$

Therefore, the volume of the solid is $\frac{\pi }{120}$