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# Washer method Let R be the region bounded by the following curves. Use the washer method to find the volume of the solid generated when R is revolved about the x-axis. $y=2 x, y=16 x^{1 / 4}$

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To find the volume of the solid generated when $R$ is revolved about $x-$ axis where the region $R$ is bounded by: $y = 2x$ and $y = 16x^{\frac{1}{4}}$ we will use $\textbf{Washer method}$:

$V = \int_a^b \pi \left( (f(x)^2 -g(x)^2 \right ) \dd{x}$

$a = 0$ and for $b$ we need to find intersection point between $y = 2x$ and $y = 16x^{\frac{1}{4}}$:

\begin{align*} 2x &= 16x^{\frac{1}{4}}\\ \frac{2x}{16x^{\frac{1}{4}}} & = 1\\ \frac{1}{8}x^{\frac{3}{4}} &= 1 \Bigg / \cdot 8 \\ x^{\frac{3}{4}} &= 8 \\ x &= 16 \\ \end{align*}

Therefore we have $a = 0$ and $b = 16$.

$f(x) = 16x^{\frac{1}{4}}$ and $g(x) = 2x$.

\begin{align*} V &= \int_a^b \pi \left(f(x)^2 - g(x)^2 \right) \dd{x} = \int_0^{16} \pi \left( (16x^\frac{1}{4})^2 - (2x)^2 \right) \dd{x} \\ &=\int_0^16 \pi(256\sqrt{x} - 4x^2)\dd{x} \\ &=256\pi\int_0^{16}\sqrt{x}\dd{x} - 4\pi \int_0^{16}x^2\dd{x} \\ &=256\pi\cdot \frac{2}{3}\eval{x^{\frac{3}{2}}}_0^{16} - 4\pi\cdot\frac{1}{3}\eval{x^3}_0^{16}\\ &=\boxed{256\pi\cdot \frac{2}{3}\cdot 64 - \frac{4\pi}{3}\cdot 16^3 = \frac{16384}{3}\pi} \end{align*}

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