<p>What is the approximate volume of the solid obtained by revolving about the x-axis the region in the first quadrant enclosed by the curves y = x^3 and y = sinx?</p>
<p>I got the answer but the answer key says its wrong. Help! Thanks.</p>
<p>Hm..... this is how I would do it:
Disc Method
1st Step: Graph sin(x) and x^3
2nd Step: Find intersection of graphs, which happens to be .928626 (x-coordinate since we are revolving around x axis) and 0
3rd Step: pi(integral from 0 to .928626) Outer radius squared minus inner radius squared (sin(x))^2- (X^3)^2 dx (This is definetly a calculator problem)</p>
<p>If you graph both functions, you can see that sinx is the "top" function, while x^3 is the "bottom" function. There intersection is at x=0.929.</p>
<p>Therefore, volume= pi*integral from 0 to 0.929 [(sinx)^2 - (x^3)^2] dx,
which should result in: 0.438025 </p>
<p>I think that's the answer, but make sure you're in the right mode (radian).</p>