Question

Surface Integrals SG(r)dA\iint_S G(\mathbf{r}) d A. Using SG(r)dA=RG(r(u,v))N(u,v)dudv\iint_S G(\mathbf{r}) d A=\iint_R G(\mathbf{r}(u, v))|\mathbf{N}(u, v)| d u d v or SG(r)dA=RG(x,y,f(x,y))1+(fx)2+(fy)2dxdy\iint_S G(\mathbf{r}) d A=\iint_{R^*} G(x, y, f(x, y)) \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2} d x d y, evaluate thise integral for the given data. (Show the details.)

G=(1+9xz)3/2,S:r=[u,v,u3],0u1,2v2G=(1+9 x z)^{3 / 2}, \quad S: \mathbf{r}=\left[\begin{array}{lll}u, & v, & u^3\end{array}\right], \quad 0 \leqq u \leqq 1,-2 \leqq v \leqq 2

Solution

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Answered 2 years ago
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We know that

SG(r)dA=RG(r(u,v))N(u,v)dudv\int\int_S G(\mathbf{r}) dA=\int\int_R G(\mathbf{r}(u,v))|\mathbf{N}(u,v)| du dv

where

N(u,v)=ru×rv\mathbf{N}(u,v)=\mathbf{r}_u\times \mathbf{r}_v

By plugging x=ux=u, y=vy=v and z=u3z=u^3 in the given function G(x,y)G(x,y) we get

G(u,v)=(1+9u4)3/2\begin{aligned} \color{#4257b2} G(u,v)=(1+9u^4)^{3/2} \end{aligned}

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