calculus 2 help?

Question

1. find the volume of the solid obtained when the areabonded by the x axis and y= -x^2+4x-3 is rotated about the line y= -1.

plz show me how to do it, cuz i really want to learn it

Answer 1

First, draw an accurate sketch of the region under consideration. The graph is a parabola which opens downward and has x-intercepts 1 and 3. That portion of the parabola between 1 and 3 is the region to be rotated about y = -1.

To set up the integral, consider a thin (width dx) rectange going from the x-axis up to the parabola. When this rectangle is taken about the line, the radius of the circular disk is approximately y+1 (because y is the height from the x-axis up to the curve, +1 more to measure down to the line y = -1). The volume of the circular disk is pi*((y + 1)^2)*dx. But now we must subtract the volume of the portion of the disk from the x-axis down to y = -1, which is pi*(1^2)*dx. Therefore our volume is pi*[(y + 1)^2 - 1]dx =pi*(y^2 + 2y)*dx.

Replace y by -x^2 + 4x - 3, and integrate from 1 to 3.

(Don't be discouraged. This is not an easy problem.)

Answer 2

y= -x^2+4x-3 is a parabola shaped like an upsidedown U that crosses the x=axis at x=1, reaches a maximum of 1 at x =2, and the goes down crossing the x-axis again at x = 3.

So the region we are looking at is that part of the parabola that lies between x = 1 and x=3 and the x-axis.

Since we are rotating the area about the line y = -1 instead of the x-axis, we must add 1 to y.

So V = pi*integ from 1 to 3 (y+1)^2dx
= pi*integ from 1 to 3 (y^2+2y+1)dx
= pi*integ 1-->3 {(-x^2+4x -3)^2 +4(-x^2+4x-3) +1}dx
So I think you can do this from here.

Answer 3

volulme
= ∫pi(-x^2+4x-3 +1)^2 dx, x from -.5858 to 3.4142
= 18.96