How do you find the volume of the region bounded #y = x²# and #y =1# is revolved about the line# y = -2#?

Question

Aurora Alvarado · Accepted Answer

See the explanation.

Here is the region (in blue) with the line #y=-2# in red.

A representative slice in black and the two radii (vertical red lines).

The volume of a washer is #(piR^2-pir^2)dx#
Where #R# is the greater and #r# the lesser radius.

In this question, #R=1-(-2) = 3# and #r=x^2-(-2) = x^2+2#

#x# goes from #-1# to #1#, so we need

#int_-1^1 pi(R^2-r^2)dx = pi int_-1^1(3^2-(x^2+2)^2)dx#

Expand the polynomial integrand and evaluate to get: #104/15pi#

Emily Ammerman · Answer

undefined To find the volume of the region bounded by $y = x^2$ and $y = 1$ when revolved about the line $y = -2$, you can use the method of cylindrical shells. The formula for the volume using cylindrical shells is:

$$ V = 2\pi \int_{a}^{b} xf(x) \, dx $$

where $f(x)$ is the function defining the outer radius, and $a$ and $b$ are the bounds of integration.

In this case, $f(x)$ is the distance from the axis of rotation ($y = -2$) to the outer curve ($y = 1$), which is $1 - (-2) = 3$.

So, $f(x) = 3$ for all $x$ in the interval $[0, 1]$.

Substituting this into the formula:

$$ V = 2\pi \int_{0}^{1} x \cdot 3 \, dx $$

Now, integrate this expression within the given bounds to find the volume of the solid.