Let R be the region enclosed by f(x) = x^2 + 2 and g(x) = (x - 2)^2. What is the volume of the solid produced by revolving R around the x-axis and then the y-axis?

Question

Christopher Agresta · Accepted Answer

Those two curves have only one point of intersection. They do not enclose a region.

graph{(y-(x^2+2))(y-(x-2)^2)=0 [-19.57, 26.06, -3.74, 19.06]} The only point of intersection is at #x=1/2#

Isabella Ackerman · Answer

undefined To find the volume of the solid produced by revolving the region enclosed by $f(x) = x^2 + 2$ and $g(x) = (x - 2)^2$ around the x-axis and then the y-axis, you can use the method of cylindrical shells.

1. First, find the points of intersection between $f(x)$ and $g(x)$ by setting them equal to each other and solving for $x$. These points determine the limits of integration.

2. Next, integrate $2\pi x \cdot (f(x) - g(x))$ with respect to $x$ over the interval determined by the points of intersection to find the volume of each shell.

3. Finally, integrate the resulting expression over the appropriate interval to find the total volume of the solid formed by revolving the region around the x-axis and then the y-axis.