Question

Find the surface area of the part of the circular paraboloid z=x2 y2 that lies inside the cylinder x2 y2=4.

Answer 1

The surface area of the circular paraboloid within the cylinder can be found using calculus, by switching to polar coordinates and computing the surface integral over the defined domain.

Step-by-step explanation:

To find the surface area of the part of the circular paraboloid z = x2 + y2 that lies inside the cylinder x2 + y2 = 4, we can use calculus, specifically multiple integrals. Since the paraboloid and the cylinder are symmetric about the z-axis, we can switch to polar coordinates (r, θ) where x = rcosθ and y = rsinθ. The radius r will range from 0 to 2, and the angle θ will vary from 0 to 2π. We will need the formula for the surface area element in polar coordinates, which for a surface defined as z = f(x, y) is dA = √(1 + (dz/dx)2 + (dz/dy)2)dxdy, or in polar coordinates, dA = √(1 + (dz/dr)2)rdrdθ.

Next, we would compute the gradient of the function in polar coordinates and then integrate it over the domain bounded by the cylinder's equation. Because this is a college-level mathematics problem, it requires familiarity with calculus, specifically with the computation of surface integrals over specified domains.

Answer 2

The part of the surface within the cylinder is defined over the domain

`R = \left\{(x,y) ~:~ x^2 + y^2 \le 4\right\}`

which we can write in polar coordinates as

`R = \left\{(r,\theta) ~:~ 0\le r\le2 \text{ and } 0 \le\theta\le2\pi\right\}`

Then the surface area of
`z=f(x,y)` over the region
`R` is given by the integral

`\displaystyle \iint_R dA = \iint_R \sqrt{1+\left((\partial f)/(\partial x)\right)^2 + \left((\partial f)/(\partial y)\right)^2} \, dx \, dy`

Let

`z = f(x,y) = x^2 + y^2`

with partial derivatives

`(\partial f)/(\partial x) = 2x`

`(\partial f)/(\partial y) = 2y`

Then in the integral, we have

`\displaystyle \iint_R dA = \iint_R √(1+4x^2+4y^2)\,dx\,dy \\\\ ~~~~~~~~ = \int_0^(2\pi) \int_0^2 r√(1+4r^2)\,dr\,d\theta \\\\ ~~~~~~~~ = 2\pi \int_0^2 r√(1 + 4r^2) \, dr \\\\ ~~~~~~~~ = \frac\pi4 \int_1^(17) \sqrt s\,ds = \boxed{\frac\pi6 \left(17^(3/2) - 1\right)}`

where we substituted
`s=1+4r^2` and
`ds=8r\,dr`.

Alternatively, we can parameterize the surface
`S` by

`\vec s(r,\theta) = r\cos(\theta)\,\vec\imath + r\sin(\theta)\,\vec\jmath + r^2\,\vec k`

with the same domain as
`R`. Then the normal vector to
`S` is

`\vec n = (\partial\vec s)/(\partial r)*(\partial\vec s)/(\partial \theta) = -2r^2\cos(\theta)\,\vec\imath - 2r^2\sin(\theta)\,\vec\jmath + r\,\vec k \\\\ \implies \|\vec n\| = √(r^2 + 4r^4) = r√(1+4r^2)`

Then the area is given by the surface integral

`\displaystyle \iint_R dS = \int_0^(2\pi) \int_0^2 \|\vec n\| \, dr \, d\theta = \int_0^(2\pi) \int_0^2 r √(1+4r^2)\, dr\,d\theta`

just as before.