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cjc0117
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Update: I figured out how to solve the problem. Nevermind.
Use a suitable change of variables to evaluate the double integral:
[itex]\int^{1}_{0}\int^{3-x}_{2x}(y-2x)e^{(x+y)^{3}}dydx[/itex]
[itex]\frac{\partial(x,y)}{\partial(u,v)}=( \frac{\partial x}{\partial u} )(\frac{\partial y}{\partial v})-(\frac{\partial x}{\partial v})(\frac{\partial y}{\partial u})[/itex]
[itex]dA=|\frac{\partial(x,y)}{\partial(u,v)}|dudv[/itex]
I made the following change of variables:
[itex]u=x+y[/itex]
[itex]v=2x[/itex]
(At first I tried u=y-2x and v=x+y but that didn't work)
Next, I solved for x and y:
[itex]x=\frac{v}{2}[/itex]
[itex]y=\frac{2u-v}{2}[/itex]
I found that the Jacobian is:
[itex]\frac{\partial(x,y)}{\partial(u,v)}=-\frac{1}{2}[/itex]
And therefore...
[itex]dA=\frac{1}{2}dudv[/itex]
Next I changed the boundaries:
[itex]x=0\rightarrow v=0[/itex]
[itex]x=1\rightarrow v=2[/itex]
[itex]y=3-x\rightarrow u=3[/itex]
[itex]y=2x\rightarrow u=\frac{3}{2}v[/itex]
Therefore, the new region S on the uv-plane is the triangle described by:
[itex]0\leq v\leq 2[/itex]
[itex]\frac{3}{2}v\leq u\leq 3[/itex]
The original integrand becomes:
[itex](y-2x)e^{(x+y)^{3}}\rightarrow \frac{2u-3v}{2}e^{u^{3}}[/itex]
So the new integral is:
[itex]\frac{1}{4}\int^{2}_{0}\int^{3}_{\frac{3}{2}v}(2u-3v)e^{u^{3}}dudv[/itex]
I used wolfram alpha to find out that the integral is equal to [itex]\frac{1}{18}(e^{27}-1)[/itex], which is the correct answer, but I am not sure how to evaluate the integral. Do I have to use another change of variables? Is there a better change of variables I could have used in the beginning?
Thank you. Any help is much appreciated.
Update:
Okay, I figured out how to solve the integral. I'll post the rest of the solution in case anybody else who has this same question ever happens across this post. What I did was I switched the order of integration so that the integral becomes:
[itex]\frac{1}{4}\int^{3}_{0}\int^{\frac{2}{3}u}_{0}(2u-3v)e^{u^{3}}dvdu[/itex]
[itex]=\frac{1}{4}\int^{3}_{0}(2uv-\frac{3}{2}v^{2})e^{u^{3}}]^{\frac{2}{3}u}_{0}du[/itex]
[itex]=\frac{1}{6}\int^{3}_{0}u^{2}e^{u^{3}}du[/itex]
Use substitution with [itex]w=u^{3}[/itex] and [itex]dw=3u^{2}du\rightarrow \frac{1}{3}dw=u^{2}du[/itex]:
[itex]\frac{1}{18}\int^{27}_{0}e^{w}dw[/itex]
[itex]=\frac{1}{18}e^{w}]^{27}_{0}[/itex]
[itex]=\frac{1}{18}(e^{27}-1)[/itex]
Homework Statement
Use a suitable change of variables to evaluate the double integral:
[itex]\int^{1}_{0}\int^{3-x}_{2x}(y-2x)e^{(x+y)^{3}}dydx[/itex]
Homework Equations
[itex]\frac{\partial(x,y)}{\partial(u,v)}=( \frac{\partial x}{\partial u} )(\frac{\partial y}{\partial v})-(\frac{\partial x}{\partial v})(\frac{\partial y}{\partial u})[/itex]
[itex]dA=|\frac{\partial(x,y)}{\partial(u,v)}|dudv[/itex]
The Attempt at a Solution
I made the following change of variables:
[itex]u=x+y[/itex]
[itex]v=2x[/itex]
(At first I tried u=y-2x and v=x+y but that didn't work)
Next, I solved for x and y:
[itex]x=\frac{v}{2}[/itex]
[itex]y=\frac{2u-v}{2}[/itex]
I found that the Jacobian is:
[itex]\frac{\partial(x,y)}{\partial(u,v)}=-\frac{1}{2}[/itex]
And therefore...
[itex]dA=\frac{1}{2}dudv[/itex]
Next I changed the boundaries:
[itex]x=0\rightarrow v=0[/itex]
[itex]x=1\rightarrow v=2[/itex]
[itex]y=3-x\rightarrow u=3[/itex]
[itex]y=2x\rightarrow u=\frac{3}{2}v[/itex]
Therefore, the new region S on the uv-plane is the triangle described by:
[itex]0\leq v\leq 2[/itex]
[itex]\frac{3}{2}v\leq u\leq 3[/itex]
The original integrand becomes:
[itex](y-2x)e^{(x+y)^{3}}\rightarrow \frac{2u-3v}{2}e^{u^{3}}[/itex]
So the new integral is:
[itex]\frac{1}{4}\int^{2}_{0}\int^{3}_{\frac{3}{2}v}(2u-3v)e^{u^{3}}dudv[/itex]
I used wolfram alpha to find out that the integral is equal to [itex]\frac{1}{18}(e^{27}-1)[/itex], which is the correct answer, but I am not sure how to evaluate the integral. Do I have to use another change of variables? Is there a better change of variables I could have used in the beginning?
Thank you. Any help is much appreciated.
Update:
Okay, I figured out how to solve the integral. I'll post the rest of the solution in case anybody else who has this same question ever happens across this post. What I did was I switched the order of integration so that the integral becomes:
[itex]\frac{1}{4}\int^{3}_{0}\int^{\frac{2}{3}u}_{0}(2u-3v)e^{u^{3}}dvdu[/itex]
[itex]=\frac{1}{4}\int^{3}_{0}(2uv-\frac{3}{2}v^{2})e^{u^{3}}]^{\frac{2}{3}u}_{0}du[/itex]
[itex]=\frac{1}{6}\int^{3}_{0}u^{2}e^{u^{3}}du[/itex]
Use substitution with [itex]w=u^{3}[/itex] and [itex]dw=3u^{2}du\rightarrow \frac{1}{3}dw=u^{2}du[/itex]:
[itex]\frac{1}{18}\int^{27}_{0}e^{w}dw[/itex]
[itex]=\frac{1}{18}e^{w}]^{27}_{0}[/itex]
[itex]=\frac{1}{18}(e^{27}-1)[/itex]
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