How do you use the disk method to find the volume of the solid formed by rotating the region bounded by #y = 2x# and #y = x^2# about the y-axis?

Question

Christopher Agresta · Accepted Answer

#y = 2x# and #y = x^2# intersect at #(0,0)# and at #(2,4)# I can't get both curves on one graph, but I'll assume you can graph the line and the parabola. Rotating around the #y# axis and using disks, means our independent variable will be #y#. The representative slice is horizontal with left end (little radius: r) on the line #x=1/2y# and right end (big radius: R) on the parabola #x=sqrty#.
The thickness of the disk will be #dy#, and the limits of integration will be #0# to #4#. #int_0^4 piR^2-pir^2 dy = piint_0^4 ((sqrty)^2-(y/2)^2) dy# #pi int_0^4 (y-y^2/4) dy = pi[y^2/2-y^3/12]_0^4 =pi[8-16/3] =(8pi)/3#

Emily Ammerman · Answer

undefined To use the disk method to find the volume of the solid formed by rotating the region bounded by $ y = 2x $ and $ y = x^2 $ about the y-axis, we integrate the area of infinitesimally thin disks along the y-axis.

The limits of integration will be the y-values where the two curves intersect. Setting $ 2x = x^2 $ gives us the intersection points. Solving this equation yields $ x = 0 $ and $ x = 2 $. Therefore, the limits of integration are from $ y = 0 $ to $ y = 4 $.

The radius of each disk is the distance from the y-axis to the curve $ y = 2x $, which is $ x $. So, the radius of the disk is $ x $.

The area of each disk is $ \pi r^2 $, where $ r $ is the radius. Substituting $ x = \sqrt{y} $ for the radius, we get $ \pi (\sqrt{y})^2 = \pi y $.

Therefore, the volume $ V $ of the solid can be expressed as:
$$ V = \int_{0}^{4} \pi y \, dy $$

Integrating this expression gives the volume of the solid formed by rotating the region bounded by $ y = 2x $ and $ y = x^2 $ about the y-axis.