What Is the Surface Area of a Sphere Inside a Paraboloid?

[jaredmt]

Homework Statement

Find the area of the sphere x^2 + y^2 + (z-2)^2 = 4 that lies inside paraboloid z = x^2 + y^2

Homework Equations

The Attempt at a Solution

when i take the equation of the spere and replace x^2 + y^2 with z i get: z(z-3) = 0
so they intersect at the plane z = 3.

were supposed to use the double integration rule. I am not sure what the parameters would be. when i convert to polar, i know feta is from 0 to 2pie. i tried making r from 0 to 3 but i got the wrong answer. idk whether i integraded the wrong equation or used the wrong parameters or both

 

[Dick]

How are we supposed to know what you did wrong if you won't tell us what you did. What did you integrate? What did you get?

 

[jaredmt]

i don't know how to type the integration symbols

the double integral was ||Rx X Ry||
where x = x, y = y, z = x^2 + y^2

so basically (skipping some steps) the integral becomes: (1 + 4x^2 + 4y^2)^.5
which becomes: r(1 + 4r^2)^.5 when put in polar form
and like i said, i tried 0<feta<2pi and 0<r<3

i got like pi/6 * some quantity. but its supposed to be 4pi

 

[Dick]

It looks like you are finding the area of the paraboloid that lies in the sphere. The question asks for the area of the sphere that lies inside the paraboloid. It's the upper cap of the sphere.

 

[Dick]

You might also notice z=r^2 on the paraboloid. The limits for z are 0->3. The limits for r are not 0->3.

 

[tiny-tim]

Hi jaredmt!

(have a theta: θ and a pi: π and a squared: ² and an integral: ∫)

when i take the equation of the spere and replace x^2 + y^2 with z i get: z(z-3) = 0
so they intersect at the plane z = 3.

That's right!

So you're now trying to find the area of a cap of a sphere of radius 2 from "height" 2 to "height" 1 (and you're told to use ∫∫).

were supposed to use the double integration rule. I am not sure what the parameters would be.

Hint: you can use either x and y parameters, or latitude and longitude parameters.

They both work (well, why wouldn't they? ), so you may as well try both of them!