How do you find the volume of the solid obtained by rotating the region bounded by the curves #y = x^3#, #x=0#, and #x=1# rotated around the #y=-2#?

Question

Scarlett Allison · Accepted Answer

The region to be rotated about y=-2 is shown here shaded in red.
Consider an element of length y and width # delta#x at a distance x from the origin. If this this is rotated about y=-2. It would form a circular disc of radius y+2. Its area would be #pi (y+2)^2#. Its volume would be #pi (y+2)^2 delta x# From this subtract the volume of coaxial region #pi 2^2 delta x#

Required volume is that of the annular region #pi (y+2)^2 delta x -4pi delta x#

=#pi(y^2 +4y)delta x#.

The volume of the solid formed by the whole region would therefore be

#int_(x=0)^1 pi (y^2 +4y) dx #. Since y is =#x^3#

Integral to solve would be #int_(x=0)^1 pi (x^6 +4x^3)dx #

= #pi[x^7 /7 +4 x^4 /4]_0^1 #
= #(8pi)/7#

William Abdalla · Answer

undefined To find the volume of the solid obtained by rotating the region bounded by the curves $y = x^3$, $x = 0$, and $x = 1$ around the line $y = -2$, you can use the method of cylindrical shells.

The formula for the volume using cylindrical shells is:
$$ V = 2\pi \int_{a}^{b} (x \cdot f(x) - h) \, dx $$

Where:
- $a$ and $b$ are the bounds of integration (in this case, $a = 0$ and $b = 1$).
- $f(x)$ is the function representing the curve ($f(x) = x^3$ in this case).
- $h$ is the distance from the axis of rotation to the curve ($h = -2$).

Substitute these values into the formula and integrate to find the volume.

Ezekiel Alexander · Answer

undefined To find the volume of the solid obtained by rotating the region bounded by the curves $ y = x^3 $, $ x = 0 $, and $ x = 1 $ around the line $ y = -2 $, we can use the method of cylindrical shells.

The formula for the volume of a solid obtained by rotating a region bounded by $ y = f(x) $, $ x = a $, $ x = b $, and rotated around the line $ y = c $ is given by:

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

In this case, $ f(x) = x^3 $, $ a = 0 $, $ b = 1 $, and $ c = -2 $.

Substituting these values into the formula, we get:

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

Now, we can expand and integrate:

$$ V = 2\pi \int_{0}^{1} (x^4 + 2x^3) \, dx $$
$$ = 2\pi \left[ \frac{x^5}{5} + \frac{2x^4}{4} \right]_{0}^{1} $$
$$ = 2\pi \left( \frac{1}{5} + \frac{1}{2} \right) $$
$$ = 2\pi \left( \frac{1}{5} + \frac{2}{5} \right) $$
$$ = 2\pi \cdot \frac{3}{5} $$
$$ = \frac{6\pi}{5} $$

Therefore, the volume of the solid obtained by rotating the region bounded by the curves $ y = x^3 $, $ x = 0 $, and $ x = 1 $ around the line $ y = -2 $ is $ \frac{6\pi}{5} $.