Answer to Question #219164 in Calculus for hajra

Question

Find the volume of the solid formed by revolving the region bounded by y=(x-2)² and y=x about the y -axis.

Answer 1

Solution;

Using the Shell Method,

Find the intersection points of the two functions,y= x and y=(x-2)²

At point of intersection,the functions are equal,

x=(x-2)²

x=x²-4x+4

x²-5x+4=0

Which is a quadratic equation,

Solving

$\frac{5_-^+\sqrt{5^2-(4×1×4)}}{2}$

x=4 and x=1

Volume by shell method;

V=$\int_a^b2πx(g(x)-f(x))$dx

V=$\int_1^42πx(x-(x-2)^2)dx$

V=$\int_1^42πx(x-(x^2-4x+4))dx$

V=$\int_1^42π(5x^2-x^3-4x)dx$

V=$[2π(\frac{5x^3}{3}-\frac{x^4}{4}-2x^2)]_1^4$

V=2π×11.243

V=70.64185

Answer:

70.64 cubic units.