How do you find the volume of a solid that is enclosed by #y=x^2-2#, #y=-2#, and #x=2# revolved about y=-2?

Question

Isabella Ackerman · Accepted Answer

#(8pi)/3# use disc method to integrate the volume: volume = #piint_a^br(x)dx# where #r(x)# is the distance from a certain point on #y=x^2-2# to the axis of rotation, #y=-2# to find a and b, find where #y=x^2-2# intersects with #y=-2#: #x^2-2=-2, x^2=0, x=0# this means the vertex of the parabola #y=x^2-2# lies on #y=-2#. one of the bounds is #x=2# as given from the problem, so the values for a and b are: #a=0# and #b=2# #r(x)# is the difference between #x^2-2# and #-2#, so #r(x)=x^2-2-(-2)=x^2# plugging in: volume = #piint_0^2x^2dx# #pi(F(2)-F(0))#, where #F(x)=1/3x^3#, or the integral of #x^2# #=pi(1/3(2)^3-1/3(0)^3)# #=pi(8/3)=(8pi)/3#

Luke Alvarez · Answer

undefined To find the volume of the solid formed by revolving the region enclosed by $ y = x^2 - 2 $, $ y = -2 $, and $ x = 2 $ about the line $ y = -2 $, you can use the method of cylindrical shells.

The volume $ V $ can be calculated using the formula:

$$ V = 2\pi \int_a^b x \cdot h(x) \, dx $$

Where $ h(x) $ represents the height of the cylinder at a given $ x $-value and $ a $ and $ b $ are the limits of integration.

In this case, the limits of integration are from $ x = -2 $ to $ x = 2 $ (the intersection points of $ y = x^2 - 2 $ and $ y = -2 $).

The height $ h(x) $ of the cylinder is the difference between the upper and lower functions at a given $ x $-value, which is $ h(x) = (x^2 - 2) - (-2) = x^2 $.

So, the integral to find the volume becomes:

$$ V = 2\pi \int_{-2}^2 x \cdot (x^2) \, dx $$

You can now integrate this expression to find the volume of the solid.