How do you find the volume of the solid #y=x^2, y=x^3# revolved about the x-axis?

Question

Emilia Aguilar · Accepted Answer

Volume # (2pi)/35 " unit^3#

graph{(y-x^2)(y-x^3)=0 [-0.1, 1.5, -0.1, 1.5]}

The Volume of Revolution about #Ox# is given by:

# V= int_(x=a)^(x=b) \ pi y^2 \ dx #

So, the volume of revolution bounded by two curves #f(x)# and #g(x)# with #f(x) gt g(x)# is given by the difference between the individual solids of revolution, thus:

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

We can see by observation that the points of intersection of the curves #y=x^2# and #y=x^3# are #(0,0)# and #(1,1)#.

So for for this problem:

# V= int_0^1 \ pi { (x^2)^2 -(x^3)^2} \ dx # # \ \ \= pi \ int_0^1 \ x^4-x^6 \ dx # # \ \ \= pi \ [x^5/5-x^7/7]_0^1# # \ \ \= pi \ { (1/5-1/7) - (0)}# # \ \ \= pi * 2/35#

# \ \ \= (2pi)/35#

Elijah Abram · Answer

undefined To find the volume of the solid generated by revolving the region between the curves $y = x^2$ and $y = x^3$ about the x-axis, you can use the method of cylindrical shells.

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

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

where:
- $ a $ and $ b $ are the x-values of the points of intersection of the curves,
- $ f(x) $ is the function representing the outer curve (in this case $ x^3 $),
- $ g(x) $ is the function representing the inner curve (in this case $ x^2 $).

First, find the points of intersection of the curves by setting $ x^2 = x^3 $, which yields $ x = 0 $ and $ x = 1 $.

So, $ a = 0 $ and $ b = 1 $.

The volume can be calculated as follows:

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

$$ = 2\pi \int_{0}^{1} \left( x^4 - x^3 \right) \, dx $$

$$ = 2\pi \left[ \frac{x^5}{5} - \frac{x^4}{4} \right]_{0}^{1} $$

$$ = 2\pi \left( \frac{1}{5} - \frac{1}{4} \right) $$

$$ = 2\pi \left( \frac{4 - 5}{20} \right) $$

$$ = 2\pi \left( -\frac{1}{20} \right) $$

$$ = -\frac{\pi}{10} $$

Thus, the volume of the solid generated by revolving the region between the curves $ y = x^2 $ and $ y = x^3 $ about the x-axis is $ -\frac{\pi}{10} $ cubic units.