Help with a solids of revolution problem. Consider the region bounded by : y = x^(1/2), x=1, and y=0. Rotate it along the y-axis, use the washer and shell method to confirm your results.

Question

Help with a solids of revolution problem.  Consider the region bounded by : y = x^(1/2), x=1, and y=0.  Rotate it along the y-axis, use the washer and shell method to confirm your results.

moonlitfate · Answer

attachment

moonlitfate · Answer

I did try to graph the function, and the areas for the bounds.  What's confusing me is figuring out what the radii are.

anonymous · Answer

Washer method for this one is really simple actually. Integrate for 0 to 1 and the formula for these problems is essentially pi*r^2 remember. So the radius in this method is simply sqrt x

anonymous · Answer

$$x^1/2 = \sqrt{x}$$ the graph look like|dw:1424125713403:dw|

anonymous · Answer

you need to make the x's--> y's since you are rotating with respect to y axis.

anonymous · Answer

$$y=\sqrt x  = x= y^2$$

anonymous · Answer

washer method is not pi r ^2. it is $$\pi (aor-oc)^2 - (aor- ic)^2$$ where aor is axis of rotation,  oc is outer curve furthest away from aor and ic is inner curve closest to aor