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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.
Thus, just why do we care regarding direction fields? Two nice pieces of information are there which can be readily determined from the direction field for a differential equation.
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How the property AM>or = GM used to get minimum value of the function......e,g for what condition of a and b does minimum value of a tan^2 x + b cot^2 x equals maximum value of a
If α, β are the zeros of the polynomial x 2 +8x +6 frame a Quadratic polynomial whose zeros are a) 1/α and 1/β b) 1+ β/α , 1+ α/β. Ans. P(x) = x 2 +8x +6 α + β = -8
4.2^2x+1-9.2^x+1=0
-2+-9
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