Answer to Question #107874 in Calculus for edward

Question

Find the volume generated by rotating the region bounded by y = x, x = 1 and y² = 4x, about the x axis.

Answer 1

Consider a region that is bounded by two curves $y=f(x)$ and $y=g(x)$ , between $x=a$ and $x=b.$

($f(x), g(x)$ are continuous and non-negative on the interval $[a,b]$ and $f(x)\leq g(x)$ )

The volume of the solid formed by revolving the region about the x-axis is

$V=\pi \int \limits_a^b\big( [f(x)]^2 -[g(x)]^2\big) dx$

We have three curves: $y=x, y^2=4x$ and $x=1$ .

The region that is bounded by them can be defined as a region that is bounded by $f(x)=2\sqrt {x}$ and $g(x)=x$ , between $x=0$ and $x=1$ .

So, the volume generated by rotating this region is:

$V=\pi \int \limits_0^1\big( (2\sqrt{x})^2 -x^2\big) dx= \pi \int \limits_0^1\big( 4x-x^2\big) dx= \pi \big(2x^2-\frac{x^3}{3}\big) \big|_0^1=\pi (2-\frac{1}{3})=\frac{5\pi}{3}$

Answer: the volume is $\frac{5\pi}{3}.$