Calculoku nº 1 (easy):?

Question

What is the volume of the solid delimited by the surfaces: z = x^2+y^2
and z = 2 - sqrt(x^2 + y^2)?

Answer 1

Oh, no, I MISREAD THIS TWICE! Back to the drawing board:

This tedious 3-D solid is made from (a) a right circular cone of height 1, vertex at z = 2, resting with its circular base, of radius 1, on top of (b) a paraboloidal volume (vertex down, at the origin) which ends in an identical base. The common base lies in the plane z = 1.

(a) Finding the volume of the right circular cone doesn't take calculus. Such a volume is given by:

V(a) = 1/3 base x height = 1/3 pi 1^2 1 = 1/3 pi.

(b) The volume of this paraboloidal volume is best found by using
(r, z) cylindrical coordinates. In those terms, this "part-volume" is given by the integral of pi r^2 dz (with r^2 = z) from z = 0 to 1. This is therefore the integral of pi z dz from z = 0 to 1, i.e. 1/2 pi.

The full volume V is then given by

V = V(a) + V(b) = 1/3 pi + 1/2 pi = 5/6 pi.

The purist would leave it there. However, I've noticed that many readers of Yahoo! Answers seem to be committed decadigitists. For them the answer is 2.618... to the number of digits exhibited.

I feel quite abject about my two earlier attempts; but at least I was self-checking and self-correcting. (Who wants to do triple integrals? Only masochists or sadists. I recommend avoiding them where possible.)

Live long and prosper.

POSTSCRIPT:

There are several mistakes in kaksi_guy's otherwise well written"solution" (Two below me). The most important is that r^2 + r - 2 = (r + 2)(r - 1). r = - 2 is an unphysical solution; in fact, the two z's clearly match at r = 1. This mistake has knock-on effects: the limits in the integrals aren't correct, and neither is the answer.

By the way, (r, z, [theta]), with r "in the plane" are CYLINDRICAL, not POLAR coordinates. If restricted to the plane, (r, theta) are indeed (plane) polar coordinates; but when you add in the z-coordinate, the name of the full set becomes cylindrical coordinates. They are still called cylindrical coordinates when it really IS a "cylindrical" problem in which the theta is only implicit, because of cylindrical symmetry. (r, z) are the traditional names for such coordinates, the word "cylindrical" itself being understoood, because of the context.

Answer 2

It’s more productive to use polar coordinates system; x=r*cos(t), y=r*sin(t); then surfaces are:
z=r^2 and z=2-r; they are easy to draw! let’s find cross-section circle:
r^2=2-r or r^2+r-2=0 or r=2 and z=1;
the whole volume I=I1+I2, where I1 is volume below the circle, I2 is volume above the notorious circle.
Elementary volume dv = S*dz, where S=pi*r^2 is the area of elementary pancake, dz is its height
I1 = integral{for r=0 until 2}of pi*r^2 * (2*r*dr) = (2pi/4) * r^4 = 8pi;
I2 = integral{for r=2 until 0} of pi*r^2 * (-dr) = (-pi/3) * r^3 = 8pi/3;
I = 8pi(1+1/3) = 32pi/3;

Answer 3

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Answer 4

A triple integral. Integrate 1 with respect to z, from x^2+y^2 to 2 - sqrt(x^2 + y^2). Integrate the result with respect to y, from -sqrt(1-x^2) to +sqrt(1-x^2) because that's the upper and lower bounds of the solid on the XY plane. (Why?) Integrate *that* result with respect to x, from -1 to +1. (Why?)

And that will be your answer.