How do you find the volume of the solid obtained by rotating the region bounded by y=5x^2 ,x=1 , and y=0, about the x-axis?

Question

Luke Alvarez · Accepted Answer

Volume # = 5pi " unit"^3#

graph{(y-5x^2)(x-1)(y)=0 [-3, 3, -2, 6]}

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

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

So for for this problem:

# V= int_0^1 \ pi (5x^2)^2 \ dx # # \ \ \= pi \ int_0^1 \ 25x^4 \ dx # # \ \ \= pi \ [(25x^5)/5]_0^1# # \ \ \= 5pi \ [x^5]_0^1# # \ \ \= 5pi (1-0)# # \ \ \= 5pi#

Christopher Agresta · Answer

undefined To find the volume of the solid obtained by rotating the region bounded by $y = 5x^2$, $x = 1$, and $y = 0$ about the x-axis, you can use the disk method.

1. Determine the limits of integration by setting up the integral with respect to $x$, from the lower bound to the upper bound, which are $x = 0$ and $x = 1$ respectively.

2. Find the radius of each disk at a given value of $x$. For this problem, the radius, $r$, is the distance from the curve $y = 5x^2$ to the x-axis, which is simply $y$.

3. Square the radius and multiply by $\pi$ to find the area of each disk.

4. Integrate the areas of the disks from $x = 0$ to $x = 1$ to find the total volume using the formula for volume by disks:

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

$$V = \int_{0}^{1} (\pi (5x^2)^2) \, dx$$

$$V = \int_{0}^{1} (25\pi x^4) \, dx$$

$$V = \pi \int_{0}^{1} (25x^4) \, dx$$

$$V = \pi \left[\frac{25x^5}{5}\right]_{0}^{1}$$

$$V = \pi \left[\frac{25}{5} - \frac{0}{5}\right]$$

$$V = \pi \cdot 5$$

$$V = 5\pi$$