How do you find the volume of the region bounded by #y=sqrt x#, y=0, x=0, and x=4 is revolved about the x-axis?

Question

Emilia Aguilar · Accepted Answer

#= 8 pi#

The volume of the small strip of width dx is given by

#dV = pi y^2 dx# where #y = sqrt x# represents the radius of the revolution at that given value for x

so #dV = pi \ x \ dx#

and

#V = pi int_0^4 \x \ dx#

# = pi[x^2/2]_0^4 = 8 pi#

Aurora Alvarado · Answer

undefined To find the volume of the region bounded by $ y = \sqrt{x} $, $ y = 0 $, $ x = 0 $, and $ x = 4 $ when revolved about the x-axis, you can use the method of cylindrical shells.

The formula for the volume of a solid generated by revolving a region bounded by $ y = f(x) $, $ y = 0 $, $ x = a $, and $ x = b $ about the x-axis is:

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

Using this formula, we can calculate the volume for the given region:

$$ V = 2\pi \int_{0}^{4} x \cdot \sqrt{x} \, dx $$

$$ V = 2\pi \int_{0}^{4} x^\frac{3}{2} \, dx $$

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

$$ V = 2\pi \left( \frac{2}{5} \cdot 4^\frac{5}{2} - \frac{2}{5} \cdot 0^\frac{5}{2} \right) $$

$$ V = 2\pi \left( \frac{2}{5} \cdot 8 - 0 \right) $$

$$ V = 2\pi \left( \frac{16}{5} \right) $$

$$ V = \frac{32}{5} \pi $$

So, the volume of the region bounded by $ y = \sqrt{x} $, $ y = 0 $, $ x = 0 $, and $ x = 4 $ when revolved about the x-axis is $ \frac{32}{5} \pi $ cubic units.