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Answer 1 · 2017-09-08T12:45:29

Volume of bounded area $= 9 π$ $=color(red)(9pi)$ cubic units

Explanation:

First let's consider this within the XY-plane
$y = \sqrt{9 - x^{2}}$ $y=sqrt(9-x^2)$
is the equation for the top half (since the root symbol restricts us to non-negative vales) of a circle with center the origin and radius $\sqrt{9} = 3$ $sqrt(9)=3$
bounding this by $y = 0$ $y=0$ and $x = 0$ $x=0$
implies we are dealing (within the XY-plane) with a quarter circle (whose radius is $3$ $3$ units)

If we rotate this about the X-axis we will obtain half of a sphere.

The volume of a sphere is $V = \frac{4}{3} π r^{3}$ $V=4/3pir^3$

Here the radius, $r = 3$ $r=3$ , and we only want half of the sphere.

So
$XXX {Volume}_{bound region} = \frac{1}{2} \times \frac{4}{3} π \times 3^{3}$ $color(white)("XXX")"Volume"_"bound region"=1/2 xx 4/3pi xx3^3$

$XXX = 9 π$ $color(white)("XXX")=9pi$ (cubic units)

Answer 2 · 2017-09-26T15:13:05

I have solved this way:

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