   Chapter 15, Problem 6P

Chapter
Section
Textbook Problem

Leonhard Euler was able to find the exact sum of the series in Problem 5. In 1736 he proved that ∑ n − 1 ∞ 1 n 2 = π 2 6 In this problem we ask you to prove this fact by evaluating the double integral in Problem 5. Start by making the change of variables x = u − v 2   y = u + v 2 This gives a rotation about the origin through the angle π / 4 . You will need to sketch the corresponding region in the uv-plane.[Hint: If, in evaluating the integral, you encounter either of the expressions (1 – sin θ)/cos θ or (cos θ)/(1 + sin θ), you might like to use the identity cos θ = sin(( π / 2 ) − θ) and the corresponding identity for sin θ.]

To determine

To evaluate: The integral 010111xydxdy (from problem 5) by using change of variables.

Explanation

Given:

The transformations are x=uv2 and y=u+v2 .

Property used: Change of Variable

Change of Variable in double integral is given by,

Rf(x,y)dA=Sf(x(u,v),y(u,v))|(x,y)(u,v)|dudv (1)

Formula used:

The Jacobian J is, (x,y)(u,v)=|xuxvyuyv|

Calculation:

Find the partial derivative of x and y with respect to u and v

If x=uv2 , then xu=12 and xv=12 .

If y=u+v2 , then yu=12 and yv=12 .

Compute the Jacobian value J.

(x,y)(u,v)=|12121212|=12(12)(12)(12)=12+12=1

The given region is in xy-plane which should converted into uv-plane

Rewrite the given transformation is,

2x=uv (2)

2y=u+v (3)

Solve the equation by performing (1)+(2) ,

2u=2(x+y)u=2(x+y)2u=(x+y)2

Solve the equation by performing (1)(2) ,

2v=2(xy)v=2(xy)2v=(xy)2v=(yx)2

From the given information uv-plane is drawn as shown below in Figure 1.

From Figure 1, the regions is {(u,v)|0u12,uvu} and the other region is {(u,v)|12u2,2+uv2u} .

Then, the integral will be,

010111xydxdy=012uududv1(u+v2)(uv2)+1222+u2ududv1(u+v2)(uv2)=012uududv1((u+v)(uv)2)+1222+u2ududv1((u+v)(uv)2)=012uududv1(u2v22)+1222+u2ududv1(u2v22)=012uu2dudv2u2+v2+1222+u2u2dudv2u2+v2

On further simplification, the value of the above integral is obtained as follows.

010111xydxdy=012uu2dudv(2u2)(1+v2(2u2))+1222+u2u2dudv(2u2)(1+v2(2u2))=012uu2dudv(2u2)(1+(v2u2)2)+1222+u2u2dudv(2u2)(1+(v2u2)2)=2012uu1(2u2)tan1(v2u2)(12u2)du+21222+u2u1(2u2)tan1(v2u2)(12u2)du=2012[12u2tan1(v2u2)]uudu+2122[12u2tan1(v2u2)]2+u2udu

On further simplification find the value of the above integral is obtained as follows.

010111xydxdy=2012[12u2(tan1(u2u2)tan1(u2u2))]du+2122[12u2(tan1(2u2u2)tan1((2u)2u2))]du=2012[12u2(tan1(u2u2)+tan1(u2u2))]du+2122[12u2(tan1(2u2u2)+tan1(2u2u2))]du

On further simplification find the value of the above integral

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