In this paper, we prove the existence of a point e in the intersection of the surfaces z x2 1 z 1 x2 y 0 and y 4. We begin by showing that e is enclosed by these surfaces. Next, we show that e is the only point in this intersection. Finally, we conclude with a brief discussion of the implications of our result.
We will be discussing the surfaces z=0(x^2+y^2-1) and z=4-y which are generated by rotating the curve about the x-axis. We will also be finding the enclosed volume between these two surfaces.
The Enclosing Surfaces
The two surfaces that enclose the region where e is located are the z=x^2-1 and z=1-x^2 surfaces. These two surfaces intersect at the point (0,0,1) and at the point (1,0,-1). The region where e is located is therefore between these two surfaces and bounded by them.
The Vector Field
In this section we will discuss the vector field which is associated with the given differential equation. We will also find the critical points of the vector field and discuss their stability.
The vector field for the given differential equation is shown below.
The method of images can be used to calculate the electric field in the region between two conducting plates. The electric field is perpendicular to the plates and is given by E 2Er where Er is the electric field due to the charge on the right plate.
To solve the problem, we first note that e is a real number since it is the sum of two squares. We now use the method of Lagrange multipliers to find the maximum and minimum values of e . We set up the function to be maximized, subject to the constraints:
e x2 1 z 1 x2 y 0
We take the partial derivatives with respect to x , y , and z and set them equal to Lagrange multiplier λ :
2 ex 2 1 z 1 x 2 y 0
Which after simplifying and solving for the roots gives us:
e 2x 3 3x 2 9x 8y 12y 4 0
When the fluid is incompressible and the surfaces z x2 1 z 1 x2 y 0 and y 4 are impermeable, the streamlines are given by the lines of intersection of these surfaces.
The result is e is enclosed by the surfaces z=x^2+1, z=1-x^2+y, and y=4.
Assuming that the surfaces given are hypersurfaces in Euclidean space, the volume enclosed by them is
where the last equality follows by observing that 1/(12x^2-1)=3/[(3x^2)(4-z)] and using the substitution u=3x^2.
In conclusion, light roasts have a light brown color and are generally preferred for milder coffee varieties. Medium roasts are medium brown in color with a stronger flavor and a non-oily surface. Medium dark roasts are rich, dark in color with some oil on the surface and with a slight bittersweet aftertaste. Dark roasts produce shiny black beans with an oily surface and a pronounced bitterness.