18.02A Topic 33: Change of variable.Read: SN: CV18.01 example: Compute I =Rx√1 − x2dx.Direct substitution: u = 1 − x2, du = −2x dx → x dx = −12du⇒ I = −12R√u du.Inverse substitution: x = sin u, dx = cos u du ⇒ I =Rsin u cos2u du.Double integration:R RRf(x, y) dA.Direct substitution: u = u(x, y), v = v(x, y).Inverse substitution: x = x(u, v), y = y(u, v).Example: x = r cos θ, y = r sin θ.Finding dA = dx dy:Jacobian =∂(x,y)∂(u,v)=xuxvyuyv,∂(u,v)∂(x,y)=uxuyvxvy.dA = dx dy =∂(x,y)∂(u,v)du dv =1|∂(u,v)∂(x,y)|du dv.(Reason soon)Change of variable forR RRf(x, y) dx dy:(A) Change f(x, y) to u, v coordinates.(B) Express dA in terms of du dv.(C) Find u, v limits on R.Example: Given the region R shown, compute I =R RR(4x2− y2)4dy dx.Change of variable: u = 2x − y, v = 2x + y.(A) 4x2− y2= uv ⇒ f(x, y) = u4v4.(B)∂(u,v)∂(x,y)=uxuyvxvy=2 −12 1= 4.⇒∂(x,y)∂(u,v)=14. ⇒ dx dy =14du dv.(C) Limits:x = 0 ⇒ u = −y, v = y ⇒ v = −u.u: -2 to 0; fix u ⇒ v: −u to 2.I =Z0u=−2Z2v=−uu4v4dv du4.(Easy to compute.)/////////12122x + y = 22x − y = 0R/////////122v = 2u = 0v = −uu = −2(continued)118.02A topic 33 2Reason for change of variables formula:Draw ’grid’ lines u = c1, v = c2(in xy-plane).∆A = area P QRS ≈ parallelogram.In a moment we will see that−−→PQ ≈ hxu∆u, yu∆ui and−→PS ≈ hxv∆v, yv∆vi.Using this we have:∆A ≈ ±xu∆u xv∆vyu∆u yv∆v= ±xuxvyuyv∆u∆v =∂(x,y)∂(u,v)∆u∆v.⇒ dA =∂(x,y)∂(u,v)du dv.Now we show−−→PQ ≈ hxu∆u, yu∆ui.Let−−→PQ = h∆x, ∆yi (see picture).The approximation formula says: ∆x ≈ xu∆u + xv∆v and ∆y ≈ yu∆u + yv∆v.Going from P to Q we have ∆v = 0 ⇒ ∆x ≈ xu∆u and ∆y ≈ yu∆u.⇒−−→PQ ≈ hxu∆u, yu∆ui.Likewise for−→PS ≈ hxv∆v, yv∆vi.PQRS∆Av = v0OOu = u0+ ∆uoov = v0+ ∆vu = u0//99ttttttttttttttPQ∆x∆yDetail from picture at leftRemark on the chain rule:∂(x, y)∂(u, v)·∂(u, v)∂(x, y)= 1 orxuxvyuyv·uxuyvxvy= I⇔ 1
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