determine-the-surface-area-of-the-portion-of-z-13-4x-2-4y-2-that-is-above-z-1-with-x-0-and-y-0-

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

determine the surface area

of the portion of z = 13 - 4 x^{2} - 4 y^{2}

that is above z = 1 with x \leq 0 and y \geq 0

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

S = \int \int_{[D]} \sqrt{1 + {(\frac{d z}{d x})}^{2} + {(\frac{d z}{d y})}^{2}} d A

z_{1} = z_{2} \Rightarrow 13 - 4 x^{2} - {4 y}^{2} = 1 \Rightarrow x^{2} + y^{2} = 3

0 ⩽ x ⩽ \sqrt{3 - y^{2}}; 0 ⩽ y ⩽ \sqrt{3}

Commented by Gbenga last updated on 14/Mar/23

t h a n k s s i r

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