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Question Number 189257 by Gbenga last updated on 14/Mar/23

$$\mathbf{determine}\mathbf{the}\mathbf{surface}\mathbf{area}$$$$\mathbf{of}\mathbf{the}\mathbf{portion}\mathbf{of}\mathbf{z}=13-4{\mathbf{x}}^2-4{\mathbf{y}}^2$$$$\mathbf{that}\mathbf{is}\mathbf{above}\mathbf{z}=1\mathbf{with}\mathbf{x}\le 0\mathbf{and}\mathbf{y}\ge 0$$

Answered by Ar Brandon last updated on 14/Mar/23

$$S=\int {\int }_{\left[D\right]}\sqrt{1+{\left(\frac{dz}{dx}\right)}^2+{\left(\frac{dz}{d\mathrm{y}}\right)}^2}dA$$$$z_1=z_2\Rightarrow 13-4x^2-{4\mathrm{y}}^2=1\Rightarrow x^2+{\mathrm{y}}^2=3$$$$0⩽x⩽\sqrt{3-{\mathrm{y}}^2};0⩽\mathrm{y}⩽\sqrt{3}$$

Commented by Gbenga last updated on 14/Mar/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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