How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #x=sqrt(y)#, x=0, y=4 about the x-axis?

Question

Elijah Abram · Accepted Answer

#(128pi)/5# cubic units.

The ends of this first-quadrant segment of the parabola #y=sqrt x# are (0, 0) and (2, 4). The shell has a paraboloid-hole in the middle. The base is circular, with radius ( range of y ) 4 units and the height is ( range of x, from 0 to 2 ) 2 units. The hole volume has to be subtracted from the volume of a right circular cylinder, of radius 4 and height 2 units.. So, the volume of the hollow solid = #pi int (4^2-y^2) d x#, with the limits, from x = 0 to x = 2 #=pi int (16-x^4) d x#, with the limits, from x = 0 to x = 2 #=pi[16x-x^5/5]#, between the limits #=(32-32/5)pi# #=(128pi)/5# cubic units.

Emma Aldridge · Answer

undefined To use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $x = \sqrt{y}$, $x = 0$, $y = 4$ about the x-axis, we integrate the volume of each cylindrical shell along the x-axis. The formula for the volume of a cylindrical shell is $V = 2\pi x h \Delta x$, where $x$ is the distance from the axis of rotation to the shell, $h$ is the height of the shell, and $\Delta x$ is the width of the shell along the x-axis.

We express $x = \sqrt{y}$ as $y = x^2$ and $y = 4$ as $x = 2$. The radius of each cylindrical shell is $x$ and the height is $4 - x^2$. We integrate from $x = 0$ to $x = 2$:

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

$$= 2\pi \int_{0}^{2} (4x - x^3) \, dx$$

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

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

$$= 2\pi \left[2(2)^2 - \frac{(2)^4}{4} - 0\right]$$

$$= 2\pi \left[8 - 4\right]$$

$$= 8\pi$$

So, the volume of the solid obtained by rotating the region about the x-axis using the method of cylindrical shells is $8\pi$.