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

`\mathbf{(\pi)/(6)[5 √(5)-1]}`

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+ ((\partial z)/(\partial x))^2+ ((\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,
`(\partial z)/(\partial x) = 2x` and
`(\partial z)/(\partial y) = 2y`

Therefore;

`S.A = \iint_D √(1+4x^2+4y^2) \ dA`

`= \iint_D √(1+4(x^2+y^2)) \ dA`

`= \int^(2 \pi)_(0) \int^(1)_(0) √(1+4r^2) \ r \ dr \d \theta`

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

`= 2 \pi * (1)/(12)[5^{(3)/(2)} - 1]`

`\mathbf{=(\pi)/(6)[5 √(5)-1]}`