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QUIZ 7, version A : MATH 251, Section 505
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Write up your result, detail your calculations if necessary and BOX your final answer
1. Let S the surface given by ~r(u, v) = hv 3 − 5u, u2 , v 2 − 3i, 0 ≤ u ≤ 1, 0 ≤ v ≤ 1.
(a) [30pts]Find the normal vector to S at the point corresponding to (u, v) = (1, 1).
(b) [35pts]Find an equation of the tangent plane to S passing through the point (−4, 1, −2).
RR
2
2
2
2. [35pts] Compute the surface integral
S xdS where S is the part of the cone z = x + y for
0 ≤ z ≤ 1 in the first quadrant.
1. (a) A normal vector to a parametric surface S is ~ru × ~rv (u, v) and r(1, 1) = (−4, 1, −2).
~ru (u, v) = h−5, 2u, 0i, so for (u, v) = (1, 1), ~ru (1, 1) = h−5, 2, 0i.
~rv (u, v) = h3v 2 , 0, 2vi, so for (u, v) = (1, 1), ~rv (1, 1) = h3, 0, 2i
~i
~ru × ~rv (1, 1) = −5
3
~j
2
0
~k 0
2
= h4, 10, −6i.
(b) The point (−4, 1, −2) corresponds to r(1, 1). Then an equation of the tangent plane to S at
(−4, 1, −2) is
~ru × ~rv (1, 1) • hx − (−4), y − 1, z − (−2)i = 4(x + 4) + 10(y − 1) − 6(z + 2) = 0 ⇔ 4x + 10y − 6z = 6
⇔ 2x + 5y − 3z = 3 .
2. We parametrize the cone by r(u, v) = hu cos v, u sin v, ui, for 0 ≤ v ≤ π2 and 0 ≤ u ≤ 1. Recall that
dS = |N (u, v)|dA where N (u, v) is a normal vector to S. We have
i
j
k sin v 1
N (u, v) = ru × rv (u, v) = cos v
−u sin v u cos v 0
= h−u cos v, −u sin v, ui.
So, |N (u, v)| =
√
2u (u ≥ 0).
Z Z
S
π
2
√
u cos v. 2udu dv
0
0
"√
#1
3
π
2u
= [sin v]02
3
0
√
2
=
.
3
Z
xdS =
Z
1
Remark : Sincepz ≥ 0, it was also possible to consider S as a graph of a function f (x, y) : S =
{(x, y, z) | z = x2 + y 2 := f (x, y), 0 ≤ z ≤ 1} but the computation of the surface integral is longer.