Question

Find the surface area of the part of the circular paraboloid z=x^2+y^2 that lies inside the cylinder X^2+y^2=1

Answer 1

Answer:

$\mathbf{\dfrac{\pi}{6}[5 \sqrt{5}-1]}$

Step-by-step explanation:

Given that:

The surface area (S.A) $z = x^2 +y^2$

Hence the S.A is of form z = f(x,y)

Then the S.A can be represented with the equation

$A(S) = \iint _D \sqrt{1+ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2} \ dA$

where :

D = cylinder $x^2 +y^2 =1$

In polar co-ordinates:

D = {(r, θ): 0≤ r ≤ 1, 0 ≤ θ ≤ 2π)

Similarly, $\dfrac{\partial z}{\partial x} = 2x$ and $\dfrac{\partial z}{\partial y} = 2y$

Therefore;

$S.A = \iint_D \sqrt{1+4x^2+4y^2} \ dA$

$= \iint_D \sqrt{1+4(x^2+y^2)} \ dA$

$= \int^{2 \pi}_{0} \int^{1}_{0} \sqrt{1+4r^2} \ r \ dr \d \theta$

$= [\theta]^{2 \pi}_{0} \dfrac{1}{8}\times \dfrac{2}{3}\begin {bmatrix} (1+4r^2)^{\dfrac{3}{2}}\end {bmatrix}^1_0$

$= 2 \pi \times \dfrac{1}{12}[5^{\dfrac{3}{2}} - 1]$

$\mathbf{=\dfrac{\pi}{6}[5 \sqrt{5}-1]}$