Answer:
68.47
Step-by-step explanation:
Given that:
The surface of the paraboloid [tex]z = 2(x^2 +y^2) \ for \ 0 \le z \le 8[/tex]
TO find the area of the given paraboloid, we need to first determine [tex]z_x \ and \ z_y[/tex].
This is because the surface area can be determined by using the formula:
[tex]\int \int _R \sqrt{z_x^2+z_y^2+ 1}\ dA[/tex]
Thus;
[tex]z_x = \dfrac{\partial }{\partial x}(2 ( x^2 +y^2) )[/tex]
[tex]z_x = 4x[/tex]
[tex]z_y = \dfrac{\partial }{\partial y }(2(x^2 +y^2))[/tex]
[tex]z_y = 4y[/tex]
Now;
[tex]\sqrt{z_x^2 + z_y^2 +1 } = \sqrt{(4x)^2 + (4y)^2+1 } \\ \\ \implies \sqrt{16x^2+16y^2+1}[/tex]
[tex]\implies \int \int \sqrt{16x^2 +16y^2+1}\ d x dy \\ \\ \implies \int \int \sqrt{16(x^2+y^2)+1 } \ dxdy[/tex]
Using the polar coordinate since the paraboloid shows the projection of a circle with [tex]x^2+y^2 =4[/tex]
[tex]\int \int \sqrt{16(x^2+y^2)+1 } \ dxdy = \int^{2 \pi}_{0} \ \int^{2}_{0} r \sqrt{16r^2+1} \ dr d\theta \\ \\ \implies \dfrac{1}{48} \Big ( ( 16r^2 +1 )^{\dfrac{3}{2}}\Big) ^2_0 \ \ \int ^{2 \pi}_{0} \ d \theta \\ \\ \implies \dfrac{2 \pi}{48} \Big ( (65^{\dfrac{3}{2}} -1 ) \Big) \\ \\ \implies \dfrac{2 \pi}{48}(523.046) \\ \\ \implies \mathbf{68. 47}[/tex]
