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Question: Find the surface area of the paraboloid $z=\frac{2}{3}[x^{\frac{3}{2}}+y^{\frac{3}{2}}]$ that lies above the $xy$-plane.

Efforts: Since we have the surface area formula $$\int\int_{D}\sqrt{1+(z_x)^2+(z_y)^2}dA$$

The region D is paraboloid lies above the $xy$-plane , I'm not getting this line because there is no restriction here.So how can calculate the limits. Thanks in advance.

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This is a valid formula to find the surface area of a solid. In this question the upper limit for z is missing in order to specify the solid. Once this is given say z = a, then the A in dA is the area of the object projected onto the xy plane.

$$\frac{2}{3}(x^{\frac{3}{2}} + y^{\frac{3}{2}}) = a$$

Hence the limit for y is from 0 to $(\frac{3a}{2} - x^{\frac{3}{2}})^{\frac{2}{3}}$

And the limit for x is from 0 to $(\frac{3a}{2})^{\frac{2}{3}}$

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The function $$z(x,y):={2\over3}\bigl(x^{3/2}+y^{3/2}\bigr)$$ is defined and nonnegative in the first quadrant $x\geq0$, $y\geq0$, and is undefined elewhere. The graph of this function is an infinite surface in ${\mathbb R}^3$, but not a paraboloid. The area of this surface is obviously $=\infty$.

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I don't know about this formula you stated. You should parameterize the paraboloid in some form g(u,v) = ( x(u,v), y(u,v), z(u,v)) then find the intervals for u and v and find the absolute value of the jacobian. In the end the surface area will be the double integral of (absolute value of the Jacobian)dudv for the intervals of u and v you found.

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