Integration application

[sxcalexc]

Just with part a.. I'm fine with obtaining an equation for the tangent.. but calculating the actual area isn't making any mathematical sense. Sure it's usually the top line minus the bottom line to work out the difference, but the integral from 1 to 3 of the tangent minus the y equation fails to yield an answer (is undefined in that domain).

Thank you!!

[pHysiX]

QU1:

y=(x-2)^0.5
dy/dx = 0.5(x-2)^-0.5

finding the tangent at (3,1):
dy/dx=0.(3-2)^-0.5
        =0.5

y-y1=m(x-x1)
tangent: y= 0.5(x-1)

Intersection: when (x-2)^0.5=0.5(x-1)
x=3

Hence, doing a sketch (sorry i can't include it here)
the area is: area of triangle of the the linear function (from 1 to 3 as your base) minus the integral from 2 to 3 of (x-1)^0.5 dx.

I think u can take it from there =]

[Mao]

b. draw the graphs, the point of intersections are (-1,-1) and (2,2). Since we are rotating it a full 360 degrees, we only need to consider the area on the right (since it will fully overlap the solid made by the area enclosed on the left). if you can visualize it, it's a parabola spun around with a cone cut out of it. More specifically, it is

, (ignoring the negative side since it is being spun a full cycle)

[Mao]

c. the formula for cylindrical shells is V = [circumference].[height].[width]

In other words,

use the method of substitution to integrate.

[sxcalexc]

Thanks, that was silly of me not to use that approach physix!! And thanks to Mao also, who gave methods which I didn't even request! Dedication, lol. In retrospect, I probably shouldn't have pasted the whole question, thanks though!

[Mao]

ahh, silly me. I didn't read that you only wanted part a.