Question

Find the volume of the solid generated when the region bounded by y=4x and y=16x is revolved about the​ x-axis. A coordinate system has an unlabeled horizontal x-axis and an unlabeled vertical y-axis. From left to right, a curve labeled y equals 16 StartRoot x EndRoot starts on the origin and rises at a decreasing rate in quadrant 1. A line labeled y equals 4 x starts at the origin and rises from left to right, intersecting the curve at the origin and in quadrant 1. The region above the line and below the curve is shaded and labeled R. A curved arrow is rotated about the x-axis. x y y=16x y=4x R

Answer 1

Answer:

The value is $V(x) = 1706.7 \pi$

Step-by-step explanation:

From the question we are told that

The first equation is $y=16\sqrt{x}$

The second equation is $y=4x$

Generally the first point of intersection of the first and second equation is x = 0

Generally the obtain the second point of intersection of the two equation we equate the two equations

So

=> $16\sqrt{x} = 4x$

=> $\sqrt{x} = \frac{4x}{16}$

=> $x = 4$

Generally the from washer method we have

$V(x) = \int\limits^4_0 {\pi [(H(x))^2 - (G(x))^2]} \, dx$

So

$H(x) = 16\sqrt{x}$

and

$G(x) = 4x$

So

$V(x) = \int\limits^4_0 {\pi [(16\sqrt{x})^2 - (4x)^2]} \, dx$

=> $V(x) = \int\limits^4_0 {\pi [256x - 16x^2]} \, dx$

=>$V(x) = \pi [256 \frac{x^2}{2} - 16 \frac{x^3}{3} ]|\left 4} \atop 0}} \right.$

=> $V(x) = \pi [256 * \frac{ 4^2}{2} - 16 * \frac{4^3}{3} ]$

=> $V(x) = 1706.7 \pi$