How do you find the volume of the region bounded by #y = (x)^(1/2)#; #y=0# and #x = 4# rotated about the x-axis?

Question

Ezra Adams · Accepted Answer

#8 pi# #y = x^(1/2)# #V = pi int_0^4y^2dx = pi/2[x^2]_0^4 = 8pi#

Isaiah Allen · Answer

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

The formula for the volume of a solid generated by rotating a curve $ y = f(x) $ from $ x = a $ to $ x = b $ about the x-axis using cylindrical shells is:

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

In this case, $ f(x) = \sqrt{x} $, $ a = 0 $, and $ b = 4 $. Substituting these values into the formula:

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

To solve this integral, simplify $ x \cdot \sqrt{x} = x^{3/2} $, then integrate:

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

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

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

$$ V = \frac{4\pi}{5}(4)^{5/2} $$

$$ V = \frac{4\pi}{5} \cdot 8\sqrt{2} $$

$$ V = \frac{32\pi\sqrt{2}}{5} $$

Therefore, the volume of the region bounded by $ y = \sqrt{x} $, $ y = 0 $, and $ x = 4 $ rotated about the x-axis is $ \frac{32\pi\sqrt{2}}{5} $.