## Introduction

In this article, we will be finding the volume of the solid enclosed by the paraboloids z4x2 y2 and z184x2 y2. We will be using the disc method and integration in order to find the volume.

## The Volume of the solid

The volume of the solid enclosed by the paraboloids z4x2 y2 and z184x2 y2 can be found by using the formula for the volume of a solid of revolution. This formula is: V=∫f(x)√1+f'(x)^2 dx, where f(x) is the function describing the shape of the solid, and f'(x) is the derivative of that function.

In this case, we are finding the volume of a solid that has been rotated around the y-axis. This means that we will need to use cylindrical coordinates in order to set up our integral. We can do this by first rewriting our functions in terms of r and θ. This gives us: z4r2 cos 2θ and z184r2 cos 2θ.

Now we can plug these into our formula for the volume of a solid of revolution to get: V=∫z4r2 cos 2θ√1+4r2 cos 2θ dθdr.

We can then solve this integral to find the volume of the solid enclosed by the paraboloids.

## The paraboloids

To find the volume of the solid enclosed by the paraboloids z4x2 y2 and z184x2 y2, we need to find the integral of these functions.

The first integral is:

∫z4x2 y2dydz=∫(z4/4)(y2+1)dydz

=∫[(1/4)(z4+4z3)y+z2]dydz

=1/4∫[(z4+4z3)y+z2]dydz

The second integral is:

∫z184x2 y2dydz=∫(9/16)(y8+1)dy⋅dz

=9/16∫[(y8/8)+1]dy⋅dz

` ...</p><br /><h2>The intersection of the paraboloids</h2><br /><p>`

To find the volume of the solid, we must first find the intersection of the paraboloids. To do this, we set z4x2 y2= z184x2 y2 and solve for z. This gives us z=14×2 y2. We can then plug this value of z into either equation to find the intersection of the paraboloids.

## The volume of the solid enclosed by the paraboloids

In order to find the volume of the solid enclosed by the paraboloids, we will need to use integral calculus. We will be integrating over the region enclosed by the paraboloids.

The first step is to set up the integral. We will be using cylindrical coordinates, so our integral will be in the form:

∫∫z4x2 y2dzdθ

The next step is to find the limits of integration. First, we need to find the intersection of the two surfaces. This can be done by setting z4=z184 and solving for x and y. This gives us: x=±y√2z. We can then plug this back into either equation to get: z=14(x2+y2).

Now that we know where the surfaces intersect, we can find our limits of integration. Our innermost limit will be 0, since we are integrating over all z values greater than 0. For our outermost limit, we will use 14(x2+y2), since this is where our surfaces intersect. For θ, we will use 0 to 2π since we are using cylindrical coordinates.

Now that we have our limits of integration, we can set up our double integral:

∫2π0∫14(x2+y2)z4x2 y2dzdθ

We can now evaluate this integral using standard methods of integration. This results in a volume of 2048π units3.