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?

Question

Emilia Aguilar · Accepted Answer

See the answer below:

Isaiah Allen · Answer

undefined 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 $.