Question 1203141
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 The area between y=x² and y+x²=8 is rotated 360° about the x-axis. 
Find the volume produced
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<pre>
The outer boundary is y = 8-x^2.

The inner boundary is y = x^2.


These two curves intersect when  

    8-x^2 = x^2,  8 = x^2+x^2,  8 = 2x^2,  x^2 = 8/2 = 4,  x = {{{sqrt(4)}}} = +/- 2.


In the interval  -2 <= x <= 2, both curve are above the x-axis.


Each vertical section  x= const  of this volume  (of this solid body)  is a RING 
with the inner radius  r= x^2  and the outer radius R = 8-x^2.


The area of this ring is  {{{pi*(R^2-r^2)}}} = {{{pi*((8-x^2)^2 - (x^2)^2)}}} = {{{pi*(64 - 16x^2)}}}.


The antiderivative is  of this function  {{{pi*(64-16x^2)}}}  is  F(x) = {{{pi*(64x - (16/3)*x^3)}}}.  


The integral from  -2  to  2  is the difference  

    F(2) - F(-2) = ({{{pi*(64*2 - (16/3)*2^3)}}}) - ({{{pi*(64*(-2) - (16/3)*(-2)^3)}}}) = {{{pi*(64*4 - 2*(16/3)*8)}}} = {{{pi*(256-256/3)}}} = {{{(2/3)*256*pi}}} = {{{(512/3)pi}}}.


Thus the desired volume is  {{{(512/3)pi}}} = {{{(512/3)*3.14159265}}} = 536.1651  (rounded).    <U>ANSWER</U>
</pre>

Solved.