# 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?

1. 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.
2. Curve and the area under the 'bottom' curve, to find the volume of a washer, we the volume of the ' curve and the volume of the ' ' curve. Ex 2 pg 463 #12 Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the given lines y = 2x 2, y = 0, x =2.

## 7.7 Volume Washer Methodap Calculus Calculator

Calculus integration geometry volume solid-of-revolution. 79 7 7 bronze badges. Finding Volume of the Solid-washer method.

#### 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.

## 7.7 Volume Washer Method Ap Calculus 14th Edition

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

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

## 7.7 Volume Washer Method Ap Calculus Frq

$= 1890 \pi$

(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
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