How do you find the volume of the solid generated by revolving the region enclosed by the parabola #y^2=4x# and the line y=x revolved about the x-axis?

Question

Emma Aldridge · Accepted Answer

The problem is equivalent to: find the volume of the solid generated by revolving the region enclosed by the parabola #y = x^2/4# and the line #x = y# both revolted about the #y# axis. So
#f_1(r)=r^2/4#
#f_2(r)=r#
#V_1(R) = 2pi int_0^R f_1(r)rdr#
#V_2(R) = 2pi int_0^R f_2(r)rdr#
#V(R) = V_2(R)-V_1(R)#
#R# is such that #f_1(r) = f_2(r)# giving #R = 4#
So #V(4) = 32pi/3#

Ezekiel Alexander · Answer

undefined To find the volume of the solid generated by revolving the region enclosed by the parabola $y^2 = 4x$ and the line $y = x$ about the x-axis, you can use the disk method.

1. First, find the points of intersection between the parabola and the line:
   $ y^2 = 4x $ and $ y = x $
   Substituting $ y = x $ into the equation of the parabola:
   $ x^2 = 4x $
   $ x^2 - 4x = 0 $
   $ x(x - 4) = 0 $
   $ x = 0 $ and $ x = 4 $
   Corresponding y-values are $ y = 0 $ and $ y = 4 $.

2. The volume $ V $ can be found using the formula for the volume of a solid of revolution using disks:
   $$ V = \pi \int_{a}^{b} [f(x)]^2 dx $$
   Where $ f(x) $ is the function describing the curve and $ [f(x)]^2 $ gives the square of the radius.

3. In this case, the radius $ r $ of each disk at a given $ x $ is the distance from the curve $ y = x $ to the x-axis, which is $ r = x $.

4. The integral for the volume becomes:
   $$ V = \pi \int_{0}^{4} x^2 dx $$

5. Integrating:
   $$ V = \pi \int_{0}^{4} x^2 dx $$
   $$ V = \pi \left[ \frac{1}{3} x^3 \right]_0^4 $$
   $$ V = \pi \left( \frac{1}{3} \cdot 4^3 - \frac{1}{3} \cdot 0^3 \right) $$
   $$ V = \pi \left( \frac{64}{3} \right) $$
   $$ V = \frac{64}{3} \pi $$