If a region is bounded by #y = sqrt(x) + 3#, #y=5#, and the y=axis and it is revolved around the y = 7, how do you find the volume?

Question

Christopher Agresta · Accepted Answer

The volume is #=120pi#

The volume of a small slice,

#dV=pi(7-(sqrtx+3))^2dx#

Therefore,

#V=int_0^4pi(7-(sqrtx+3))^2dx#

#V=piint_0^4 (49-(x+6sqrtx+9))dx#

#=piint_0^4 (49-x-6sqrtx-9)dx#

#=pi[40x-x^2/2-6x^(3/2)/(3/2)]_0^4#

#=pi(160-8-32)=120pi#

graph{(y-(sqrtx+3))(y-5)(y-7)=0 [-5.335, 10.465, 2.226, 10.126]}

Luke Alvarez · Answer

undefined To find the volume of the solid obtained by revolving the region bounded by $ y = \sqrt{x} + 3 $, $ y = 5 $, and the $ y$-axis around the line $ y = 7 $, we use the method of cylindrical shells.

1. Determine the limits of integration. Since the region is bounded by $ y = \sqrt{x} + 3 $ and $ y = 5 $, we need to find the intersection points of these two curves. Set $ \sqrt{x} + 3 = 5 $ and solve for $ x $ to find the upper limit of integration.

$$
\sqrt{x} + 3 = 5
$$

$$
\sqrt{x} = 5 - 3 = 2
$$

$$
x = 4
$$

So, the upper limit of integration is $ x = 4 $. The lower limit of integration is $ x = 0 $ since the region is bounded by the $ y$-axis.

2. Setup the integral. The volume $ V $ of the solid can be calculated using the formula for the volume of cylindrical shells:

$$
V = \int_{0}^{4} 2\pi rh \, dx
$$

where $ r $ is the distance from the axis of revolution to the shell, and $ h $ is the height of the shell.

3. Express $ r $ and $ h $ in terms of $ x $. Since we are revolving around $ y = 7 $, the distance from $ y = \sqrt{x} + 3 $ (or $ y = 5 $) to $ y = 7 $ is $ 7 - (\sqrt{x} + 3) $. Therefore, $ r = 7 - y $.

The height $ h $ of each cylindrical shell is $ \sqrt{x} + 3 - 5 $, which simplifies to $ \sqrt{x} - 2 $.

4. Substitute $ r $ and $ h $ into the integral and solve:

$$
V = \int_{0}^{4} 2\pi (7 - y)(\sqrt{x} - 2) \, dx
$$

5. Evaluate the integral to find the volume $ V $.