How do you find the volume of a solid that is enclosed by #y=x+1#, #y=x^3+1#, x=0 and y=0 revolved about the x axis?

Question

Christopher Agresta · Accepted Answer

The volume #=(3pi)/4# A mall volume (rotation about the x axis #dV=piy^2dx# We have to find the volume enclosed by the curves so #dV=piy_1^2dx-piy_2^2dx# Hence #dV=pi(x^3+1)^2dx-pi(x+1)^2dx# #dV=pi((x^6+2x^3+1)-(x^2+2x+1))dx# #dV=pi(x^6+2x^3-x^2-2x)dx# and the limits of the integral when #y=0# #0=x+1#=>#x=-1# and #x=0# So #V=piint_-1^0(x^6+2x^3-x^2-2x)dx# #V=pi(x^7/7+(2x^4)/4-x^3/3-(2x^2)/2)_-1^0# #V=pi(0-(-1/7+1/2+1/3-1))= pi(1/7+1-5/6)# #V=pi*13/42=(3pi)/4#

Emilia Aguilar · Answer

undefined To find the volume of the solid generated by revolving the region bounded by the curves $y = x + 1$, $y = x^3 + 1$, $x = 0$, and $y = 0$ about the x-axis, you can use the method of cylindrical shells. The volume $V$ can be calculated using the integral:

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

Where $a$ and $b$ are the x-values of the points of intersection of the curves, and $f(x)$ and $g(x)$ represent the functions that define the upper and lower boundaries of the region, respectively.

First, we find the intersection points of the curves:
$$ x + 1 = x^3 + 1 $$
$$ x^3 - x = 0 $$
$$ x(x^2 - 1) = 0 $$
$$ x(x + 1)(x - 1) = 0 $$

So, the intersection points are $x = -1$, $x = 0$, and $x = 1$.

Next, we integrate from 0 to 1, as the region of interest lies between those x-values:
$$ V = \int_{0}^{1} 2\pi x \cdot ((x^3 + 1) - (x + 1)) \, dx $$

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

$$ V = \int_{0}^{1} 2\pi (x^4 - x^2) \, dx $$

$$ V = 2\pi \left[\frac{x^5}{5} - \frac{x^3}{3}\right]_{0}^{1} $$

$$ V = 2\pi \left(\frac{1}{5} - \frac{1}{3}\right) $$

$$ V = \frac{2\pi}{15} $$

So, the volume of the solid is $\frac{2\pi}{15}$ cubic units.