surface area - 2

[logistic_guy]

Find the surface area of the given surface. The portion of the paraboloid $\displaystyle z = x^2 + y^2$ below the plane $\displaystyle z = 4$.

 

[khansaheb]

Find the surface area of the given surface. The portion of the paraboloid $\displaystyle z = x^2 + y^2$ below the plane $\displaystyle z = 4$.

show us your effort/s to solve this problem.

 

[logistic_guy]

Method 3


Let $\displaystyle f(x,y) = z = x^2 + y^2$

$\displaystyle \iint\limits_S \ dS = \int\int \sqrt{(f_x)^2 + (f_y)^2 + 1} \ dx \ dy$

$\displaystyle = \int_{-2}^{2}\int_{-\sqrt{4 - y^2}}^{\sqrt{4 - y^2}} \sqrt{4x^2 + 4y^2 + 1}\ dx \ dy$

$\displaystyle = \int_{0}^{2\pi}\int_{0}^{2} r\sqrt{4r^2 + 1} \ dr \ d\theta$

$\displaystyle = \frac{1}{8}\int_{0}^{2\pi}\int_{1}^{17} \sqrt{u} \ du \ d\theta$

$\displaystyle = \frac{1}{8}\int_{0}^{2\pi} \frac{2}{3}u^{3/2}\bigg|_{1}^{17}\ d\theta$

$\displaystyle = \frac{1}{8} \frac{2}{3}(17^{3/2} - 1)2\pi$

$\displaystyle = \frac{(17^{3/2} - 1)\pi}{6} \approx 36.18$

 

[StarRider]

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