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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=2x,y=16x1/4y=2 x, y=16 x^{1 / 4}

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To find the volume of the solid generated when RR is revolved about xx- axis where the region RR is bounded by: y=2xy = 2x and y=16x14y = 16x^{\frac{1}{4}} we will use Washer method\textbf{Washer method}:

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

a=0a = 0 and for bb we need to find intersection point between y=2xy = 2x and y=16x14y = 16x^{\frac{1}{4}}:

2x=16x142x16x14=118x34=1/8x34=8x=16\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=0a = 0 and b=16b = 16.

f(x)=16x14f(x) = 16x^{\frac{1}{4}} and g(x)=2xg(x) = 2x.

V=abπ(f(x)2g(x)2) ⁣dx=016π((16x14)2(2x)2) ⁣dx=016π(256x4x2) ⁣dx=256π016x ⁣dx4π016x2 ⁣dx=256π23x320164π13x3016=256π23644π3163=163843π\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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