Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Section 15.5 - Surface Area - 15.5 Exercise - Page 1028: 10

Answer

The area of the surface $z$ is $$ A(S) =4 \pi . $$
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Work Step by Step

$$ f(x,y,z)=x^{2}+y^{2}+z^{2}=4 , $$ when $z=1$ we get $$ x^{2}+y^{2}=3. $$ so, $$ D=\left\{ (x, y) | x^{2}+y^{2} \leq 3\right\}. $$ and $$ z=\sqrt {4-x^{2}-y^{2}}, $$ $$ f_{x}=\frac{-x}{\sqrt {4-x^{2}-y^{2}}} , \quad f_{y}=\frac{-y}{\sqrt {4-x^{2}-y^{2}}} $$ Thus $$ \begin{aligned} A(S) &=\\ &\iint_{D} \sqrt{\left[(-x)\left(4-x^{2}-y^{2}\right)^{-1 / 2}\right]^{2}+\left[(-y)\left(4-x^{2}-y^{2}\right)^{-1 / 2}\right]^{2}+1} d A \\ &=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \sqrt{\frac{r^{2}}{4-r^{2}}+1} r d r d \theta \\ &=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \sqrt{\frac{r^{2}+4-r^{2}}{4-r^{2}}} r d r d \theta \\ &=\int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \frac{2 r}{\sqrt{4-r^{2}}} d r d \theta \\ &\left.=\int_{0}^{2 \pi}\left[-2\left(4-r^{2}\right)^{1 / 2}\right]_{r=0}^{r=\sqrt{3}} d \theta \\ =\int_{0}^{2 \pi}(-2+4) d \theta=2 \theta\right]_{0}^{2 \pi}=4 \pi \end{aligned} $$ Thus, the area of the surface $z$ is $$ A(S) =4 \pi . $$
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