Let R be the region in the first quadrant bounded by the graph of $y=8-x^(3/2)$ , the x-axis, and the y-axis. What is the best approximation of the volume of the solid generated when R is revolved about the x-axis?

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Answer 1 · 2018-01-20T01:27:05

The volume of the solid generated when R is revolved about the $x$ -axis is $848/15pi$ or approximately $177.6$ .

The area of a circle with a radius equal to $8 - x^(3/2)$ at any given $x$ is $pi*(8 - x^(3/2))^2$ .

$pi * (8 - x^(3/2))^2$

$pi * (64 - 16x^(3/2) + x^3)$

$pix^3 - pi16x^(3/2) + 64pi$

We wish to figure out exactly what region we're looking at. This can be done by looking at the $x$ -intercept of $f(x)$ .

$8 - x^(3/2) = 0$

$8 = x^(3/2)$

$8^(2/3)=x$

$4 = x$

It has an $x$ -intercept of $4$ .

With this, we know to integrate $pix^3 - pi16x^(3/2) + 64pi$ from $0$ to $4$ .

The antiderivative of $pix^3 - pi16x^(3/2) + 64pi$ is:

$pi/3 color(red)x^2 - (32pi)/5 color(red)x^(5/2) + 64pi color(red)x$

So,

$int_0^4 (pix^3 - pi16x^(3/2) + 64pi)dx$

$= pi/3 color(red)4^2 - (32pi)/5 color(red)4^(5/2) + 64pi color(red)4 - (pi/3 color(red)0^2 - (32pi)/5 color(red)0^(5/2) + 64pi color(red)0)$

$= pi/3 color(red)4^2 - (32pi)/5 color(red)4^(5/2) + 64pi color(red)4$

$= pi*(1/3 color(red)4^2 - (32)/5 color(red)4^(5/2) + 64* color(red)4)$

$= pi * (16 / 3 - 1024/5 + 256)$

$= pi * (80 / 15 - 3072 / 15 + 3840 / 15)$

$= 848/15pi$

$~~ 177.6$

Let R be the region in the first quadrant bounded by the graph of y=8-x^(3/2)y=8-x^(3/2), the x-axis, and the y-axis. What is the best approximation of the volume of the solid generated when R is revolved about the x-axis?