calculus using volume and integrals?

Question

Let Q be the region bounded by the graph of the parabola y=1-x^2 and the x-axis. There is a line y=c that divides Q into two regions of equal volume when Q is rotated around the x-axis. Find the value of c.

does anyone know how i would go about solving this?? any help would be wonderful! thanks!!

Answer 1

First your region is bounded by the parabola to be between x=1 and x=-1. At both these values the curve crosses the y axis. So your limits on x ar from -1 to 1 and the limits on y are from the x axis (y=0) to the parabola y=1-x^2.

You could set this up as a triple integral with a very simple volume elements but my preference is to use a disk for a volume element. The disk has radius y=1-x^2 and thickness dx giving a differential volume of .π(1-x^2)^2 dx. Integrating for the volume (S for integral):

S(from x=-1 to +1) π(1-x^2)^2 dx = π((x^5)/5 - 2(x^3)/3 + x)

Putting in the limits: π(2/5 - 4/3 + 2) = π 16/15

Now on to part 2. You need to find a y-limit of c that gives half the volume. This means the volume must be π 8/15.

The problem really comes down to figuring out the limits for the c boundary. It all comes down to the fact that changing c changes the x limits as well as the y. Of course, the parabola defines this. For any value of c. the limits on x are defined by:

c = 1 - x^2

Solving for x: x = + sqrt(1-c), to x = -sqrt(1-c)

The integral is like before except that the differential volume is now a washer with inner radius of c so the differential volume is π((1-x^2)^2 - c^2) dx.

The integral then is:

S(from x=-sqrt(1-c) to sqrt(1-c)) π((1-x^2)^2 - c^2) dx

The integration is a bit of a nuisance but the result is:

(2π/3)sqrt(1-c)(2 - 3c^2 + c)

This equals half the volume giving the relation for c of:

π 8/15 =(8π/15)sqrt(1-c)(2 - 3c^2 + c)

Reducing:

1 = sqrt(1-c)(2 - 3c^2 + c)

Unfortunately, this does not have a closed form solution for c but can be shown numerically to be 0.58637

Answer 2

This is kind of poorly worded, but I think I understand what is needed. The line y=c will also be rotated which makes a cylinder that divides the rotated volume into two parts of equal volume.


Integral from c to 1 of 2*sqrt(1-y) dy = Integral from 0 to c of 2*sqrt(1-y) dy

Solve for c

Approximate Answer c = 0.37004

Answer 3

before everything, because we are integrating with understand to x, we want x to be in [0, a]. So, the quantity interior the 1st quadrant is then equivalent to ?(x = 0 to a) ?b^2 (a million - x^2/a^2) dx = ?b^2 * ?(x = 0 to a) (a million - (a million/a^2) x^2) dx, **noting that a and b are constants** = ?b^2 * [x - (a million/a^2) * (a million/3)x^3] {for x = 0 to a} = ?b^2 * [a - (a million/a^2) * (a million/3)a^3] = ?b^2 * (2/3)a = (2/3)?ab^2. So, the quantity of the excellent ellipsoid equals 2 * (2/3)?ab^2 = (4/3)?ab^2. i wish this helps!