Change of Variables In Multiple Integrals

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Transcription:

Change of Variables In Multiple Integrals

When we convert a double integral from rectangular to polar coordinates, recall the changes that must be made to, and da. = (, r θ = rcosθ θ = (, r θ = rsinθ da = r drdθ r

In the polar coordinate sstem, an element of area is generall a rectangle corresponding to a range of values for r and a range for θ. θ c θ d a r b r

However, in the coordinate sstem, this rectangle usuall takes on a different shape, and the formula for an element of area changes. ΔA r Δr Δθ θ r

his leads to the following formula for the double integral in polar coordinates. θ f (, da = f ( r cos θ, r sin θ r drdθ r

We ll now develop a general method for finding change of variable formulas such as this one. θ f (, da = f ( r cos θ, r sin θ r drdθ r

Suppose we have a rectangle in an st-coordinate sstem and a pair of functions that converts (s,t coordinates into (, coordinates. = st (, t = st (, s

Suppose also that these functions are differentiable and that the transformation from the st-coordinates to -coordinates is one-to-one. = st (, t = st (, s

hen because of differentiabilit, local linearit will be present and a small rectangle in the st sstem will be mapped onto approimatel a parallelogram in the -coordinate sstem. t = = st (, st (, s

If we add some coordinates, then it looks like this. t ( st, +Δt = = ( st, ( s+δs, t st (, s st (, ( ( s, t+ Δ t, ( s, t+δt b a ( ( s, t, ( s, t ( ( s+δ s, t, ( s+δs, t

An element of area in our -coordinate sstem is represented b a parallelogram. t ( st, +Δt = = ( st, ( s+δs, t st (, s st (, ( ( s, t+ Δ t, ( s, t+δt b a ( ( s, t, ( s, t ( ( s+δ s, t, ( s+δs, t

he area of this parallelogram is given b the norm of a cross product. Area = a b ( ( s, t+ Δ t, ( s, t+δt b a ( ( s, t, ( s, t ( ( s+δ s, t, ( s+δs, t

a = s+δs t s t i + s+δs t st ˆj Notice that, ( ( (,, ˆ ( (, (, ˆ Δ si + Δs ˆj s s b= st+δt st i+ st+δt st ˆj ( (, (, ˆ (, (, ˆ Δ ti + Δt ˆj t t ( ( ( st, +Δ t, ( st, +Δt b a ( ( st,, ( st, ( ( s+δ s, t, ( s+δs, t

Hence, iˆ ˆj kˆ a b Δs Δ s 0 = s t s t kˆ Δ Δ Δ Δ s s s t s t Δt Δt 0 t t ( ( st, +Δ t, ( st, +Δt b a ( ( st,, ( st, ( ( s+δ s, t, ( s+δs, t

And, a b Δs Δt Δs Δ t = ΔsΔt s t s t s t s t ( ( st, +Δ t, ( st, +Δt b a ( ( st,, ( st, ( ( s+δ s, t, ( s+δs, t

he epression inside the last absolute value sign is called the Jacobian, and it is usuall written as follows: (, s t = = (, s t s t s t s t ( ( st, +Δ t, ( st, +Δt b a ( ( st,, ( st, ( ( s+δ s, t, ( s+δs, t

herefore, (, (, Area =Δ A = a b ΔsΔ t = ΔsΔt s t s t s t ( ( st, +Δ t, ( st, +Δt b a ( ( st,, ( st, ( ( s+δ s, t, ( s+δs, t

And as a result of local linearit, this approimation improves as the changes in s and t become smaller. (, (, Area =Δ A = a b ΔsΔ t = ΔsΔt s t s t s t ( ( st, +Δ t, ( st, +Δt b a ( ( st,, ( st, ( ( s+δ s, t, ( s+δs, t

And finall, f(, da= lim f(, ΔA ( ( st, +Δ t, ( st, +Δt b a ( ( st,, ( st, ( ( s+δ s, t, ( s+δs, t ΔA 0 (, ( st, = lim f((,, s t (, s t ΔsΔt Δs, Δt 0 (, ( st, = f ( st (,, ( s, t dsdt

Now let s verif that this formula works for polar coordinates. f(, da= lim f(, ΔA ΔA 0 ( ( st ( ( st,, = lim f (, ΔsΔ t = f (, dsdt Δs, Δt 0,, ( ( st, +Δ t, ( st, +Δt b a ( ( st,, ( st, ( ( s+δ s, t, ( s+δs, t

(, ( r, θ f (, da = f ( st (,, ( st, drdθ = rcosθ = rsinθ (, r θ cosθ r sinθ = = (, r θ sinθ rcosθ r θ ( 2 2 2 2 = rcos θ + rsin θ = r cos θ + sin θ = r (, (, r θ = = r = r

herefore, (, ( r, θ f (, da= f ( rcos θ, rsin θ drdθ = f( rcos θ, rsin θ rdrdθ

Let s tr one more eample! Consider the equation below for an ellipse. a 2 2 + = b 2 2 1

he following substitution will change it into an equation for a unit circle. = a s = b t 2 2 2 2 ( a s ( b t 2 2 + = 1 + = 1 s + t = 1 2 2 2 2 a b a b

Now find the absolute value of the Jacobian. = a s = b t (, s t a 0 = = = (, st 0 b s t ab (, (, st = ab

We can now easil find the area of the ellipse. Area of ellipse= da = ab dsdt ellipse unit circle = ab dsdt = ab π = π ab unit circle

We can now easil find the area of the ellipse. Area of ellipse= da = ab dsdt ellipse unit circle = ab dsdt = ab π = π ab unit circle B the wa, another wa to arrive at the same result is to notice that: = a s d= ads dd = ab dsdt = b t d = bdt

If we have a function of three variables, then our Jacobian looks like this. s t u (, z, = (,, stu s t u z z z s t u

Find the volume of the ellipsoid. 2 2 2 + + = 1 2 2 2 z a b c

he coordinate change below transforms the ellipsoid into a unit sphere. = a s = b t z = c u 2 2 2 + + = 1 2 2 2 z a b c s t u a 0 0 ( z,, = = 0 b 0 = (,, stu s t u 0 0 c z z z s t u abc

2 2 2 + + = 1 2 2 2 z a b c ( z,, Volume= dv = dsdtdu V (,, stu = abc dsdtdu = abc dsdtdu π 4 4 = abc = 3 3 π abc

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