16.6
Parametric Surfaces and Their Areas
Problem #22, page 1079
Find a parametric representation for the part of the elliptic paraboloid
x + y2 + 2z2 = 4 that lies in front of the plane x = 0.
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It is easy to solve the equation for the elliptic paraboloid for x in terms of y and z,
so an easy parametric representation is:
x = 4 – y2 – 2z2, y = y, z = z. [note: since x > 0, y2 + 2z2 < 4]
Then, r(y,z) = (4 – y2 – 2z2) i + y j + z k
If you would prefer to express your answer in terms of the variables u and v, then
r(u,v) = (4 – u2 – 2v2) i + u j + v k
[u2 + 2v2 < 4]
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For some applications you may want to use a different parametric representation.
Let u = t 2 cosθ , v = t sinθ , 0 ≤ t < 2
Then 0 ≤ u2 + 2v2 = 2t2 < 4
r(t,θ) = (4 – 2t2) i + t 2 cosθ j + tsinθ k,
0≤t< 2
Problem 26, page 1079
Find a parametric representation for the part of the plane z = x + 3 that lies inside
the cylinder x2 + y2 = 1.
An example similar to problems 33 – 36, page 1080
Find an equation of the tangent plane to the given parametric surface at the
specified point.
x = u2, y = u – v2,
z = v2;
(1,0,1)
First of all notice that for the point (1,0,1):
x = 1 → u2 = 1. y = 0 → u – v2 = 0, z = 0 → v2 = 1,
and so we get u = 1, v = ±1.
r(u,v) = ⟨ u2, u – v2, v2 ⟩
We can find a normal to the plane by crossing ru and rv.
ru(u,v) =
rv(u,v) =
ru x rv =
at u = 1, v = 1, this becomes ⟨ 2, -4, -4 ⟩
at u = 1, v = -1 we have ⟨ -2, 4, 4 ⟩
Let n = ⟨ -2, 4, 4 ⟩ and the equation for the plane is:
-2x + 4y + 4z = (-2 + 0 + 4) which cleans up nicely to:
-x + 2y + 2z = 1
Problem #44, page 1080
Find the surface area of the part of the paraboloid x = y2 + z2 that lies inside the
cylinder y2 + z2 = 9.
An example not from the text, but similar to problem #58 on page 1080.
Find the surface area of that part of the cylinder x2 + z2 = a2 that lies inside the
cylinder x2 + y2 = a2.