surface integral and volume integral?

Question

surface area and volume of the region defined by (x-1)^2+y^2+z^2<4 and (x+1)^2+y^2+z^2<4.

im pretty sure the region is an ellipsoid, and i am having trouble with the parameterization

Answer 1

Not an ellipsoid at all actually, but the intersection of two open balls of radius 2, one centered at (1, 0, 0) and the other at (-1, 0, 0). Note that this is a symmetric region, so you can represent the volume and surface area as twice the volume of one half of the region, say {(x, y, z): (x-1)²+y²+z²<4 ∧ x<0}. To find the volume of that half, note that x must lie between -1 and 0. Moving (x-1)² to the other side, we see that y² + z² < 4-(x-1)², so -√(4-(x-1)²) < y < √(4-(x-1)²). Finally, moving y to the other side, we see that -√(4-(x-1)²-y²) < z < √(4-(x-1)²-y²). So for the volume of the entire region (which is twice the volume of the half-region just described), we obtain:

2 [-1, 0]∫ [-√(4-(x-1)²), √(4-(x-1)²)]∫ [-√(4-(x-1)²-y²), √(4-(x-1)²-y²)]∫1 dz dy dx.

Of course, this integral is difficult to evaluate as written, so we shall change to cylindrical coordinates, with the cylinder aligned on the x-axis, rather than the z-axis as typical. So y=r cos θ and z=r sin θ. This gives us the more pleasant integral:

2 [-1, 0]∫ [0, 2π]∫ [0, √(4-(x-1)²)]∫r dr dθ dx

Integrating the inside integral and evaluating:

2 [-1, 0]∫ [0, 2π]∫ (4-(x-1)²)/2 dθ dx

And again:

2π [-1, 0]∫ 4-(x-1)² dx

And finally:

2π (4x-(x-1)³/3)|[-1, 0]
2π ((0-(-1)³/3) - (-4-(-2)³/3))
2π (1/3 + 4 - 8/3)
10π/3

So the volume of the region is 10π/3.

The surface area of the region is the portion of the left sphere that lies inside the right sphere plus the portion of the right sphere that lies inside the left. Note that like the volume, this is also a symmetric region, so you may also simply find the surface area of the left sphere segment and then double it to find the entire surface area of the region. Now, we can parameterize the surface area by noting that it occurs on the boundary of the sphere, so that (x-1)²+y²+z²=4. This in turn means that x=1-√(4-y²-z²) (we know to take the negative sign, because we are dealing with the left half of the region). So we can parameterize using r(u, v) = (1-√(4-u²-v²), u, v). Now, we must find the limits of u and v. Since we are dealing only with the left half, we want x to be less than zero, so √(4-u²-v²)>1, which means u²+v²<3. In other words, u and v are bounded by the circle centered at the origin with radius √3. This makes sense, because if you look at where the left half of the surface joins with the right half, it is the circle centered at the origin, lying entirely within the yz-plane, and having radius √3. Now, we recall the formula for a surface integral:

[{u²+v²<3}]∬1 |r_u×r_v| dA

Now we need to compute r_u and r_v:

∂r/∂u = (u/√(4-u²-v²), 1, 0)
∂r/∂v = (v/√(4-u²-v²), 0, 1)

Now compute |r_u×r_v|. We don't actually have to find r_u×r_v (thankfully), since we know that |a×b| = √(a·a*b·b - (a·b)²). Using this formula:

r_u·r_u = u²/(4-u²-v²) + 1
r_v·r_v = v²/(4-u²-v²) + 1
r_u·r_v = uv/(4-u²-v²)

|r_u×r_v|² = (u²/(4-u²-v²) + 1)(v²/(4-u²-v²) + 1) - (uv/(4-u²-v²))²
|r_u×r_v|² = u²v²/(4-u²-v²)² + u²/(4-u²-v²) + v²/(4-u²-v²) + 1 - u²v²/(4-u²-v²)²
|r_u×r_v|² = u²/(4-u²-v²) + v²/(4-u²-v²) + 1
|r_u×r_v|² = u²/(4-u²-v²) + v²/(4-u²-v²) + (4-u²-v²)/(4-u²-v²)
|r_u×r_v|² = 4/(4-u²-v²)
|r_u×r_v| = 2/√(4-u²-v²)

Finally, we write the final integral, remembering to double the integral since we are only integrating over the left half of the area:

2 [-√3, √3]∫ [-√(3-v²), √(3-v²)]∫ 2/√(4-u²-v²) du dv

Of course, this integral is difficult to evaluate as written, so we make a change of variables to polar coordinates. In this case, we obtain the simpler integral:

2 [0, 2π]∫ [0, √3] ∫ 2/√(4-r²) r dr dθ

And this integral may be evaluated directly. Doing so:

2 [0, 2π]∫ -2√(4-r²) |[0, √3] dθ
2 [0, 2π]∫ -2(1-2) dθ
4 [0, 2π]∫1 dθ
8π

Thus the surface area is 8π. And we are done. Hallelujah!

Answer 2

Line Integrals are used in vector calculus and physics often to calculate the work done on a particle and for similar physical applications. You will find all three of these integrals used in electromagnetism a lot where you need to find work done over an electric field.