# 7.7 Volume Washer Methodap Calculus

How do you find the volume of the region bounded by #y=6x# #y=x# and #y=18# is revolved about the y axis? Calculus Applications of Definite Integrals Determining the Volume of a Solid of Revolution 1 Answer. Kuta Software - Infinite Calculus Name Volumes of Revolution - Washers and Disks Date Period For each problem, find the volume of the solid that results when the region enclosed by the curves is revolved about the the x-axis. 1) y = −x2 + 1 y = 0 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4 −2 2 4 6 8. $begingroup$ The washer method is a mnemonic for setting up a correct integral when finding the volume of a solid of revolution. Once you have an integral (which you seem to have here), the washer method has done its job, and all that's left is pure calculation. $endgroup$ – Arthur Aug 24 '19 at 10:12. 5.2 volumes: disks and washers 401 for all x so we can instead compute the volume with a single integral: V = Z 2 0 p h x2 1 i 2 dx = p Z 2 0 x4 2x2 +1 i dx = p 1 5 x5 2 3 x3 + x 2 0 = p 32 5 16 3 +2 = 46p 15 or about 9.63. Find the volume of the solid formed by revolving the region between f(x) = 3 x and the horizontal line y = 2.

### How do you find the volume of the region bounded by #y=6x# #y=x# and #y=18# is revolved about the y axis? ## 7.7 Volume Washer Methodap Calculus Solver

#### Explanation:

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.

## Calculus Volume Washer Method

Rewrite the region: $x = \frac{1}{6} y$, $x = y$ and $y = 18$

As $y$ goes from $0$ to $18$, x goes from $x = \frac{1}{6} y$ on the left, to $x = y$ on the right.
The greater radius is $R = y$ and the lesser is $r = \frac{1}{6} y$

Evaluate
$\pi {\int}_{0}^{18} \left({R}^{2} - {r}^{2}\right) \mathrm{dy} = \pi {\int}_{0}^{18} \left({y}^{2} - {\left(\frac{y}{6}\right)}^{2}\right) \mathrm{dy}$

$= \frac{35 \pi}{36} {\int}_{0}^{18} {y}^{2} \mathrm{dy}$

$= 1890 \pi$

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

## Related topic

Determining the Volume of a Solid of Revolution
Questions

## 7.7 Volume Washer Methodap Calculus Problems 