It's not clear whether you're talking about the bounded region to the left or right of the <em>y</em>-axis, but due to symmetry it doesn't really matter.
The two curves <em>y</em> = 8 and <em>y</em> = 8<em>x</em> ² intersect for
8 = 8<em>x</em> ² → <em>x</em> ² = 1 → <em>x</em> = -1, <em>x</em> = 1
but we only care about one of these; take the positive solution, so the region of interest lies to the right of the <em>y</em>-axis.
Each washer has an outer radius (distance from the "top" curve to the axis of revolution) of 8 and an inner radius (distance from the "bottom" curve) of 8<em>x</em> ². Take the heights to be some small length ∆<em>x</em> . Then each has a volume of <em>π</em> (outer radius)² - <em>π</em> (inner radius)², or
<em>π</em> (64 - 64<em>x</em> ⁴) ∆<em>x</em>
Now take <em>n</em> such washers, take their sum, and let <em>n</em> → ∞ and ∆<em>x</em> → 0. This amounts to computing the integral,
∫₀¹ <em>π</em> (64 - 64<em>x</em> ⁴) d<em>x</em> = <em>π</em> (64<em>x</em> - 64/5 <em>x</em> ⁵) |₀¹
… = <em>π</em> (64 - 64/5)
… = 256<em>π</em>/5