Answer :
Answer:
[tex]\mathbf{\dfrac{\pi}{6}[5 \sqrt{5}-1]}[/tex]
Step-by-step explanation:
Given that:
The surface area (S.A) [tex]z = x^2 +y^2[/tex]
Hence the S.A is of form z = f(x,y)
Then the S.A can be represented with the equation
[tex]A(S) = \iint _D \sqrt{1+ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2} \ dA[/tex]
where :
D = cylinder [tex]x^2 +y^2 =1[/tex]
In polar co-ordinates:
D = {(r, θ): 0≤ r ≤ 1, 0 ≤ θ ≤ 2π)
Similarly, [tex]\dfrac{\partial z}{\partial x} = 2x[/tex] and [tex]\dfrac{\partial z}{\partial y} = 2y[/tex]
Therefore;
[tex]S.A = \iint_D \sqrt{1+4x^2+4y^2} \ dA[/tex]
[tex]= \iint_D \sqrt{1+4(x^2+y^2)} \ dA[/tex]
[tex]= \int^{2 \pi}_{0} \int^{1}_{0} \sqrt{1+4r^2} \ r \ dr \d \theta[/tex]
[tex]= [\theta]^{2 \pi}_{0} \dfrac{1}{8}\times \dfrac{2}{3}\begin {bmatrix} (1+4r^2)^{\dfrac{3}{2}}\end {bmatrix}^1_0[/tex]
[tex]= 2 \pi \times \dfrac{1}{12}[5^{\dfrac{3}{2}} - 1][/tex]
[tex]\mathbf{=\dfrac{\pi}{6}[5 \sqrt{5}-1]}[/tex]