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The area of ​​the portion of the sphere $ x^{2} + y^{2} +z^{2} = 1$ located inside of the cylinder $x = x^{2} + y^{2}$, and above the plane $z = 0$.

I'm stuck, so any tip will be helpful

Thanks in advance!

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1 Answer 1

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Hint:

$$z^2 = 1-x^2-y^2$$

$$S = \int\int_{D} \sqrt{1+f_{x}^2 +f_{y}^2}dA$$

$$ f_x = -\frac{x}{z}, f_y = -\frac{y}{z}, dA = rdrd\theta$$

$$S = \int_{0}^{\pi}\int_{0}^{cos(\theta)} \frac{1}{\sqrt{1-r^2}}rdrd\theta$$

Evaluate this and that will be your answer.

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  • $\begingroup$ I should get 2 as answer? $\endgroup$
    – user78723
    Commented Aug 11, 2014 at 15:09
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    $\begingroup$ You should get an answer $\pi-2$ $\endgroup$ Commented Aug 12, 2014 at 6:59

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