Answer to Question #129172 in Calculus for Jade

Question

Multiple choice: Find the volume of the solid that results when the region enclosed by the graphs of y=x^2 and y=2x is rotated about the x-axis

A:4/15 B:1/6 C:2pi/3 D:64pi/15

Answer 1

Answer:

$Volume = \dfrac {64π} {15} \space units^3$

Solution

We are given functions: $y = 2x \space, and \space y = x^2$

Solving for limits of x:

$x^2 = 2x \\ => x^2 - 2x = 0\\ => x(x-2) = 0 \\ => either \space x = 0 \space or \space x -2 = 0$

$\therefore \space x = 0 \space or \space x = 2$

When a region is rotated about the $x-axis$, Volume is given by:

$V = π \int_{x_1}^{x_2} y^2 dx$

Now,

$When \space y = 2x, \space y^2 = (2x)^2$

$=> y^2 = 4x^2$ , and

$When \space y = x^2, \space y^2 = (x^2)^2$

$=> y^2 = x^4$

$\therefore \space Volume = π \int_{0}^{2} (4x^2-x^4)dx$

$= π \Big \lceil \dfrac {4x^3}{3} - \dfrac {x^5}{5} \Big \rceil_{0}^{2}$

$= π \Big \lceil \Big(\dfrac {4(2^3)}{3} - \dfrac {(2^5)} {5} \Big) - \Big(\dfrac {4(0^3)}{3} - \dfrac {(0^5)} {5} \Big) \Big \rceil$

$= π \Big \lceil \Big(\dfrac {4(8)}{3} - \dfrac {32} {5} \Big) - \Big(\dfrac {4(0)}{3} - \dfrac {(0)} {5} \Big) \Big \rceil$

$= π \Big \lceil \Big(\dfrac {32}{3} - \dfrac {32} {5} \Big) - \Big(0 - 0 \Big) \Big \rceil$

$= π \Big \lceil \dfrac {32}{3} - \dfrac {32} {5} \Big \rceil$

$= π \Big \lceil \dfrac {160-96}{15} \Big \rceil$

$= π \Big( \dfrac {64}{15} \Big)$

$= \dfrac {64π}{15} \space units^3$