<p>Find the volume of the curve by the cyndrical shell method bounded by this:
y=sqrt(1-x^2)
y=x
x-axis</p>
<p>revolved around y-axis.</p>
<p>I'm confused how you even set this up!</p>
<p>i beleive you set the radius equal to x, and the highth equal to y then plug into the equation. draw it out first to make shure it makes sense because im doing this from my head.</p>
<p>wouldnt it just be:</p>
<p>(2 pi)*(integral from 0 to 1 of [(x)(sqrt(1-x^2)]</p>
<p>b/c height is y (or sqrt 1-x^2) & length x is x.</p>
<p>so i guess u'd have to do a let u thing.</p>
<p>is that enough info?</p>
<p>i think thats right?</p>
<p>If I understand your question correctly, you subtract one function from the dominant function in order to get the area between. You're half-way there.</p>
<p>It is revolved around the y axis though.</p>
<p>Don't you have to make it equal to x?</p>
<p>Shell's method: You must ensure that the equations given are written in terms of the other variable of the axis of revolution.
If given an equation in terms of x, and told to revolve it around y-axis, write the integral in terms of x and dx. This means that if an equation is given in terms of either x or y, and you are told to revolve it around the opposite variable (y=8-x^2 around y axis), you are fine, write it in terms of x. </p>
<p>BUT if you are given an equation in terms of y, and told to revolve it around y axis, solve that equation and write in terms of x, then plug it into the formula.</p>