How do you find the volume bounded by #y=sqrtx# and the lines y=0 and x=4 revolved about the y=-1?

Question

Aurora Alvarado · Accepted Answer

# (56 pi )/3 #

the volume of a small element is given by

#Delta V = pi ((1+sqrt(x))^2 - (1)^2 ) Delta x#
# = pi (1+2 sqrt(x) + x - 1 ) Delta x#
# = pi (2 sqrt(x) + x ) Delta x#

So # V = pi \int_0^4 \ 2 sqrt(x) + x \ dx#

#= pi [4/3 x^{3/2} + x^2/2 ]_0^4#

#= pi( [4/3 * 8 + 8 ]- 0)#

# = (56 pi )/3 #

Ezekiel Alexander · Answer

undefined To find the volume bounded by $ y = \sqrt{x} $ and the lines $ y = 0 $ and $ x = 4 $ revolved about the line $ y = -1 $, you use the method of cylindrical shells.

1. Determine the limits of integration. Since the region is revolved around the line $ y = -1 $, and the boundaries are $ y = 0 $ and $ x = 4 $, the limits of integration for $ y $ are $ 0 $ to $ 2 $ (since $ y = \sqrt{x} $ intersects $ x = 4 $ at $ y = 2 $).

2. Set up the integral. The volume $ V $ is given by $ V = \int_{a}^{b} 2\pi rh \, dy $, where $ r $ is the distance from the axis of revolution to the curve, and $ h $ is the height of the shell.

3. Express $ r $ and $ h $ in terms of $ y $. Since we're revolving around the line $ y = -1 $, $ r = y + 1 $ and $ h = 4 - y^2 $ (distance between $ x = 4 $ and $ y = \sqrt{x} $).

4. Substitute $ r $ and $ h $ into the integral and integrate with respect to $ y $ from $ 0 $ to $ 2 $:

$$ V = \int_{0}^{2} 2\pi(y + 1)(4 - y^2) \, dy $$

5. Compute the integral to find the volume.