How do you find the volume of the region bounded by #y=6x# #y=x# and #y=18# is revolved about the y axis?

Question

Cameron Alexander · Accepted Answer

Use washers. to get #V = 1890 pi#

The region is the bounded region in:

graph{(y-6x)(y-x)(y-0.0001x-18) sqrt(81-(x-9)^2)sqrt(85-(y-9)^2)/sqrt(81-(x-9)^2)sqrt(85-(y-9)^2) = 0 [-28.96, 44.06, -7.7, 28.83]}

Taking vertical slices and integrating over #x# would require two integrals, so take horizontal slices.

Rewrite the region: #x=1/6y#, #x=y# and #y=18#

As #y# goes from #0# to #18#, x goes from #x = 1/6y# on the left, to #x=y# on the right. The greater radius is #R = y# and the lesser is #r = 1/6y#

Evaluate #pi int_0^18 (R^2-r^2) dy = pi int_0^18 (y^2 - (y/6)^2) dy#

# = (35pi)/36 int_0^18 y^2 dy#

# = 1890pi#

(Steps omitted because once it is set up, I think this is a straightforward integration.)

Axel Alvarez · Answer

undefined To find the volume of the region bounded by the curves $y = 6x$, $y = x$, and $y = 18$ revolved about the y-axis, you can use the method of cylindrical shells.

The limits of integration will be determined by the points where the curves intersect. Setting $6x = x$ to find the intersection point:

$$6x = x$$  
$$5x = 0$$

So, $x = 0$ is the intersection point.

The radius $r$ of the cylindrical shell at $x$ is $x$, because the axis of rotation is the y-axis. The height $h$ of the cylindrical shell at $x$ is the difference between the $y$-values of the upper and lower curves, which is $18 - 6x$ - $x = 18 - 7x$.

The volume $V$ of each cylindrical shell is given by $V = 2\pi rh$.

Therefore, the total volume is given by the integral:

$$V = \int_{0}^{6} 2\pi x(18 - 7x) \, dx$$