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Surface Area with Polar Coordinates
We will be searching for at surface area in polar coordinates in this part. Note though that all we're going to do is illustrate the formulas for the surface area as most of these integrals tend to be quite difficult.
We want to locate the surface area of the region found through rotating,
r = f (θ)
α < θ < β
about the x or y-axis.
Like we did in the tangent and arc length sections we will write the curve in terms of a set of parametric equations.
x= r cosθ
= f (θ) cos θ
y = r sin θ
= f (θ) sin θ
If we now make use of the parametric formula for finding the surface area we'll obtain,
S = ∫ 2Πy ds rotation about x-axis
S = ∫ 2Πx ds rotation about y-axis
Where
ds = √r2 + (dr/dθ)2 dθ
r = f (θ) , α < θ < β
Note: since we will pick up a dθ from the ds we'll require to substitute one of the parametric equations in for x or y depending upon the axis of rotation. This will frequently mean that the integrals will be rather unpleasant.
use the simplex method to solve the following lp problem. max z = 107x1 + x2 + 2x3 subject to 14x1 + x2 - 6x3 + 3x4 = 7 16x1 + x2 - 6x3 3x1 - x2 - x3 x1,x2,x3,x4 > = 0
We have to enclose a field along with a fence. We contain 500 feet of fencing material & a building is on one side of the field & thus won't require any fencing. Find out the dime
sin2A+cos2A
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