How do you find the volume of a solid that is enclosed by #y=3x^2# and y=2x+1 revolved about the x axis?

Question

Scarlett Allison · Accepted Answer

#V=pi(4/3+2+1-9/5-4/3(-1/27)-2(1/9)-(-1/3)+1/5(-1/27))=pi1088/405# Let's find the intersections between the line and the parabola so that we have to solve #3x^2-2x-1=0# this gives two solutions #x_1=-1/3# and #x_2=1#. From the first Guldino theorem, the sought volume is given by the integral #2piint_(-1/3)^1dxint_(3x^2)^(2x+1)ydy# that can be rewritten as #2piint_(-1/3)^1(2x+1)^2/2-(3x^2)^2/2dx# Rewrite after developing the binomial square so that it gets #piint_(-1/3)^1(4x^2+4x+1-9x^4)dx# This solved gets #pi(4/3x^3+2x^2+x-9/5x^5)|_(-1/3)^1#

Paisley Johnson · Answer

undefined To find the volume of the solid enclosed by $ y = 3x^2 $ and $ y = 2x + 1 $ revolved about the x-axis, you can use the method of cylindrical shells. Follow these steps:

1. Determine the limits of integration by finding the points of intersection between the two curves $ y = 3x^2 $ and $ y = 2x + 1 $.
2. Set up the integral for the volume using the formula for cylindrical shells: $ V = \int_{a}^{b} 2\pi x f(x) \, dx $, where $ f(x) $ represents the height of the shell at each x-value, and $ a $ and $ b $ are the limits of integration.
3. Calculate the height of each cylindrical shell at a given x-value by finding the difference between the y-values of the upper and lower curves.
4. Integrate the expression from step 2 with respect to $ x $ over the determined limits of integration.
5. Evaluate the integral to find the volume of the solid.

These steps will give you the volume of the solid formed by revolving the region between the curves $ y = 3x^2 $ and $ y = 2x + 1 $ about the x-axis.