Question

21. A surface S is defined by the vector function $r(u,v) = < u, v, 2u >$. Find an expression for $\iint_S \vec{F} \cdot \vec{n}dS$, where $\vec{n}$ is the upper unit normal, $0 \le u \le 1$, and $0 \le v \le 1$. a. $\int_0^1 \int_0^1 \vec{F} \cdot < -2, 0, 1 > du dv$ b. $\int_0^1 \int_0^1 \vec{F} \cdot < 2, 0, 1 > du dv$ c. $\int_0^1 \int_0^1 \vec{F} \cdot < 0, 2, 1 > du dv$ d. $\int_0^1 \int_0^1 \vec{F} \cdot < 0, -2, 1 > du dv$ e. none of these

          21. A surface S is defined by the vector function $r(u,v) = < u, v, 2u >$. Find an expression for
$\iint_S \vec{F} \cdot \vec{n}dS$, where $\vec{n}$ is the upper unit normal, $0 \le u \le 1$, and $0 \le v \le 1$.
a. $\int_0^1 \int_0^1 \vec{F} \cdot < -2, 0, 1 > du dv$
b. $\int_0^1 \int_0^1 \vec{F} \cdot < 2, 0, 1 > du dv$
c. $\int_0^1 \int_0^1 \vec{F} \cdot < 0, 2, 1 > du dv$
d. $\int_0^1 \int_0^1 \vec{F} \cdot < 0, -2, 1 > du dv$
e. none of these

21. A surface S is defined by the vector function r(u,v) = < u, v, 2u >. Find an expression for
F⃗·n⃗dS, where n⃗ is the upper unit normal, 0 ≤ u ≤ 1, and 0 ≤ v ≤ 1.
a. ∫0^1 ∫0^1 F⃗· < -2, 0, 1 > du dv
b. ∫0^1 ∫0^1 F⃗· < 2, 0, 1 > du dv
c. ∫0^1 ∫0^1 F⃗· < 0, 2, 1 > du dv
d. ∫0^1 ∫0^1 F⃗· < 0, -2, 1 > du dv
e. none of these

Added by Valerie W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Babita Kumari

Step 1

The unit normal vector is given by the cross product of the partial derivatives of r with respect to u and v: n = (dr/du) x (dr/dv) dr/du = <1, 2uv, u> dr/dv = <0, u^2, 0> Taking the cross product: n = <2uv, -u, u^2> Show more…

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A surface S is defined by the vector function r(u=< uv^2u >. Find an expression for âˆ¬S FÂ·dS, where n is the unit normal, 0â‰¤uâ‰¤1, and 0â‰¤vâ‰¤1. a. F=<-2,0,1>du dv b. F=<2,0,1>du dv c. F=<0,2,1>du dv d. F=<0,-2,1>du dv e. None of these