# How do you find the volume of a solid obtained by revolving the graph of #y=9x*sqrt(16-x^2)# over [0,16] about the y-axis?

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To find the volume of a solid obtained by revolving the graph of ( y = 9x\sqrt{16-x^2} ) over the interval ([0,16]) about the y-axis, we use the method of cylindrical shells. The formula for finding the volume using cylindrical shells is ( V = 2\pi \int_{a}^{b} xf(x) dx ), where ( f(x) ) represents the height of the function, and ( a ) and ( b ) represent the interval over which the function is being revolved.

Substituting ( f(x) = 9x\sqrt{16-x^2} ) into the formula, we have:

[ V = 2\pi \int_{0}^{16} x \cdot 9x\sqrt{16-x^2} dx ]

We simplify the integrand and perform the integration:

[ V = 18\pi \int_{0}^{16} x^2\sqrt{16-x^2} dx ]

This integral can be evaluated using trigonometric substitution. Let ( x = 4\sin(\theta) ), then ( dx = 4\cos(\theta)d\theta ), and ( \sqrt{16-x^2} = 4\cos(\theta) ). Substituting these into the integral, we get:

[ V = 18\pi \int_{0}^{\frac{\pi}{2}} (4\sin(\theta))^2 \cdot (4\cos(\theta)) \cdot 4\cos(\theta) d\theta ]

[ V = 18\pi \int_{0}^{\frac{\pi}{2}} 64\sin^2(\theta)\cos^2(\theta) d\theta ]

Using the double-angle identity ( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} ), we rewrite the integral as:

[ V = 18\pi \int_{0}^{\frac{\pi}{2}} 64\left(\frac{1 - \cos(2\theta)}{2}\right)\cos^2(\theta) d\theta ]

[ V = 18\pi \int_{0}^{\frac{\pi}{2}} 32\cos^2(\theta) - 32\cos^2(2\theta) d\theta ]

[ V = 18\pi \left[\frac{32}{3}\sin(\theta) - \frac{32}{3}\sin(2\theta)\right]_{0}^{\frac{\pi}{2}} ]

[ V = 18\pi \left(\frac{32}{3} - 0\right) ]

[ V = \frac{1536}{3}\pi ]

[ \boxed{V = 512\pi} ]

Therefore, the volume of the solid obtained by revolving the graph of ( y = 9x\sqrt{16-x^2} ) over the interval ([0,16]) about the y-axis is ( 512\pi ).

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