Answer:
Step-by-step explanation:
From the given information:
The domain D of integration in polar coordinates can be represented by:
D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;
The partial derivates for z = xy can be expressed as:
[tex]y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}[/tex]
Thus, the area of the surface is as follows:
[tex]\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA[/tex]
[tex]= \iint_D \sqrt{x^2 +y^2 +1 } \ dA[/tex]
[tex]= \int^{2 \pi}_{0} \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr \ d \theta[/tex]
[tex]=2 \pi \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr[/tex]
[tex]= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0[/tex]
[tex]= 2 \pi \times \dfrac{1}{3} \Bigg [ (37)^{3/2} - 1 \Bigg][/tex]
[tex]= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ][/tex]
