# How do you find the volume of the solid generated by revolving the region bounded by the graphs y=x, y=0, y=4, x=6, about the line x=6?

May 30, 2017

$V = \frac{208}{3} \pi$

#### Explanation:

Here is the region described:

We can find this area in one of two ways. Method 1 uses intuitive geometric properties to find the volume, while Method 2 uses the Disk Method to find the volume.

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Method 1

Rotating this shape about $x = 6$ will produce a conical frutsum.

In other words, it will be the volume of the cone with base radius $6$ and height $6$ (as defined by the lines $y = x$ and $x = 6$) MINUS the volume of the cone with base radius $2$ and height $2$ (as defined by the area between $y = x , y = 4 , \mathmr{and} x = 6$).

Therefore, as described above, the volume is:

$V = \frac{1}{3} \pi {\left(6\right)}^{2} \left(6\right) - \frac{1}{3} \pi {\left(2\right)}^{2} \left(2\right)$

$V = \frac{1}{3} \pi \left(216\right) - \frac{1}{3} \pi \left(8\right)$

$V = \frac{208}{3} \pi$

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Method 2

Since the axis of rotation is vertical, we will integrate with respect to $y$.

The Disk Method formula, therefore, is:

$V = {\int}_{a}^{b} \pi {\left(x - \text{axis}\right)}^{2} \mathrm{dy}$

$V = \pi {\int}_{0}^{4} {\left(x - 6\right)}^{2} \mathrm{dy}$

And since we can make the substitution $y = x$,

$V = \pi {\int}_{0}^{4} {\left(y - 6\right)}^{2} \mathrm{dy}$

$V = \pi \cdot {\left[{\left(y - 6\right)}^{3} / 3\right]}_{0}^{4}$

$V = \pi \left[{\left(- 2\right)}^{3} / 3 - {\left(- 6\right)}^{3} / 3\right]$

$V = \pi \left[- \frac{8}{3} + \frac{216}{3}\right]$

$V = \frac{208}{3} \pi$