logistic_guy
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Find the surface area of the given surface. The portion of the paraboloid \(\displaystyle z = x^2 + y^2\) below the plane \(\displaystyle z = 4\).
surface area - 2
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show us your effort/s to solve this problem.
logistic_guy
Full Member
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- Apr 17, 2024
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Method 3
Let \(\displaystyle f(x,y) = z = x^2 + y^2\)
\(\displaystyle \iint\limits_S \ dS = \int\int \sqrt{(f_x)^2 + (f_y)^2 + 1} \ dx \ dy\)
\(\displaystyle = \int_{-2}^{2}\int_{-\sqrt{4 - y^2}}^{\sqrt{4 - y^2}} \sqrt{4x^2 + 4y^2 + 1}\ dx \ dy\)
\(\displaystyle = \int_{0}^{2\pi}\int_{0}^{2} r\sqrt{4r^2 + 1} \ dr \ d\theta \)
\(\displaystyle = \frac{1}{8}\int_{0}^{2\pi}\int_{1}^{17} \sqrt{u} \ du \ d\theta \)
\(\displaystyle = \frac{1}{8}\int_{0}^{2\pi} \frac{2}{3}u^{3/2}\bigg|_{1}^{17}\ d\theta \)
\(\displaystyle = \frac{1}{8} \frac{2}{3}(17^{3/2} - 1)2\pi \)
\(\displaystyle = \frac{(17^{3/2} - 1)\pi}{6} \approx 36.18\)
Let \(\displaystyle f(x,y) = z = x^2 + y^2\)
\(\displaystyle \iint\limits_S \ dS = \int\int \sqrt{(f_x)^2 + (f_y)^2 + 1} \ dx \ dy\)
\(\displaystyle = \int_{-2}^{2}\int_{-\sqrt{4 - y^2}}^{\sqrt{4 - y^2}} \sqrt{4x^2 + 4y^2 + 1}\ dx \ dy\)
\(\displaystyle = \int_{0}^{2\pi}\int_{0}^{2} r\sqrt{4r^2 + 1} \ dr \ d\theta \)
\(\displaystyle = \frac{1}{8}\int_{0}^{2\pi}\int_{1}^{17} \sqrt{u} \ du \ d\theta \)
\(\displaystyle = \frac{1}{8}\int_{0}^{2\pi} \frac{2}{3}u^{3/2}\bigg|_{1}^{17}\ d\theta \)
\(\displaystyle = \frac{1}{8} \frac{2}{3}(17^{3/2} - 1)2\pi \)
\(\displaystyle = \frac{(17^{3/2} - 1)\pi}{6} \approx 36.18\)