Question

find the area of the surface. the part of the sphere x2 y2 z2 = 4z that lies inside the paraboloid z = x2 y2.

Answer 1

To find the area of the surface that lies inside the paraboloid and the sphere, we need to determine the intersection between the two surfaces. By setting the equations of the paraboloid and the sphere equal to each other and simplifying, we find the equation of the intersection. Then, we use the formula for the surface area of a sphere to find the area of the surface.

Step-by-step explanation:

To find the area of the surface that lies inside the paraboloid and the sphere, we need to determine the intersection between the two surfaces.

First, let's find the intersection equation by setting the equations of the paraboloid and the sphere equal to each other:

x2 + y2 + z2 = 4z and z = x2 + y2

Substituting the second equation into the first equation, we get:

x2 + y2 + (x2 + y2)2 = 4(x2 + y2)

Simplifying this equation, we have:

x2 + y2 + x4 + 2x2y2 + y4 = 4x2 + 4y2

Rearranging terms, we get:

x4 + 2x2y2 + y4 - 3x2 - 3y2 = 0

Now, we use the formula for the surface area of a sphere to find the area of the surface:

Surface area = 4πr2

In this case, the radius of the sphere is 2 because the equation is x2 + y2 + z2 = 4z. So, the surface area of the sphere is 4π(2)2 = 16π.

Answer 2

The surface area of the part of the sphere that lies inside the paraboloid is simply
`\(1\)` square unit.

To find the surface area of the part of the sphere
`$x^2 + y^2 + z^2 = 4z$` that lies inside the paraboloid
`$z = x^2 + y^2$`, we need to set up a double integral over the region of intersection between these two surfaces.

The surface area can be calculated using the following formula:

`\[S = \iint_R \sqrt{1 + \left((\partial z)/(\partial x)\right)^2 + \left((\partial z)/(\partial y)\right)^2} \, dA\]`

Where:

• 
  `\(R\)` is the region of intersection between the sphere and the paraboloid in the
  `\(xy\)`-plane.
• 
  `\((\partial z)/(\partial x)\)` and
  `\((\partial z)/(\partial y)\)` are the partial derivatives of
  `\(z\)` with respect to
  `\(x\)` and
  `\(y\)`.
• 
  `\(dA\)` is an infinitesimal area element in the
  `\(xy\)`-plane.

Let's begin by finding the partial derivatives of
`\(z\)` with respect to
`\(x\)` and
`\(y\)`:

Given:

1. Sphere:
`\(x^2 + y^2 + z^2 = 4z\)`

2. Paraboloid:
`\(z = x^2 + y^2\)`

We'll first find
`\((\partial z)/(\partial x)\) and \((\partial z)/(\partial y)\)`:

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

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

Now, we need to find the region of intersection
`\(R\)` in the
`\(xy\)`-plane by solving for
`\(x\)` and
`\(y\)` in both equations. We'll set the two equations equal to each other:

`\(x^2 + y^2 + z^2 = 4z\)` (Sphere)

`\(z = x^2 + y^2\)` (Paraboloid)

Substitute the second equation into the first:

`\(x^2 + y^2 + (x^2 + y^2) = 4(x^2 + y^2)\)`

Now, simplify and solve for
`\(x^2 + y^2\)`:

`\((x^2 + y^2) + (x^2 + y^2) - 4(x^2 + y^2) = 0\)`

Combine like terms:

`\(-2(x^2 + y^2) = 0\)`

Divide by -2:

`\(x^2 + y^2 = 0\)`

This equation represents a single point at the origin
`\((0, 0)\)` in the
`\(xy\)`-plane. So, the region of intersection
`\(R\)` is just the point
`\((0, 0)\)`.

Now that we have
`\((\partial z)/(\partial x)\)`,
`\((\partial z)/(\partial y)\)`, and
`\(R\)`, we can calculate the surface area using the formula:

`\[S = \iint_R \sqrt{1 + \left((\partial z)/(\partial x)\right)^2 + \left((\partial z)/(\partial y)\right)^2} \, dA\]`

Since
`\(R\)` is just a single point, the integral reduces to:

`\[S = √(1 + (2x)^2 + (2y)^2) \cdot dA\]`

At the point
`\((0, 0)\)`, this simplifies to:

`\[S = √(1 + 0 + 0) \cdot dA = 1 \cdot dA\]`

So, The answer is
`\(1\)` square unit.