Evaluate
[MATH]\iint_A e^{3x+7y} dxdy[/MATH]
Where [MATH]A=\{(x,y) \in \mathbb{R}^2 \ | \ (3x+7y)^2 \leq x-2y <8\}[/MATH]
I've tried this substitution
[MATH]r=x-2y[/MATH][MATH]s=3x+7y[/MATH]
Then it is [MATH]x=\frac{7}{13}r+\frac{2}{13}s[/MATH] and [MATH]y=-\frac{3}{13}r+\frac{1}{13}s[/MATH], hence
[MATH]|\det J(r,s)|=\left| \det \begin{pmatrix} \frac{7}{13} & \frac{2}{13} \\ -\frac{3}{13} & \frac{1}{13}\end{pmatrix}\right|=1[/MATH]
So
[MATH]\iint_A e^{3x+7y} dxdy=\iint_B e^s drds[/MATH]
With [MATH]B=\{(r,s) \in \mathbb{R}^2 \ | \ s^2 \leq r <8\}[/MATH]. So it is
[MATH]\iint_B e^s drds=\int_{-2\sqrt{2}}^{2\sqrt{2}} \left(\int_{s^2}^8 e^s dr\right)ds[/MATH]
Is this correct?
[MATH]\iint_A e^{3x+7y} dxdy[/MATH]
Where [MATH]A=\{(x,y) \in \mathbb{R}^2 \ | \ (3x+7y)^2 \leq x-2y <8\}[/MATH]
I've tried this substitution
[MATH]r=x-2y[/MATH][MATH]s=3x+7y[/MATH]
Then it is [MATH]x=\frac{7}{13}r+\frac{2}{13}s[/MATH] and [MATH]y=-\frac{3}{13}r+\frac{1}{13}s[/MATH], hence
[MATH]|\det J(r,s)|=\left| \det \begin{pmatrix} \frac{7}{13} & \frac{2}{13} \\ -\frac{3}{13} & \frac{1}{13}\end{pmatrix}\right|=1[/MATH]
So
[MATH]\iint_A e^{3x+7y} dxdy=\iint_B e^s drds[/MATH]
With [MATH]B=\{(r,s) \in \mathbb{R}^2 \ | \ s^2 \leq r <8\}[/MATH]. So it is
[MATH]\iint_B e^s drds=\int_{-2\sqrt{2}}^{2\sqrt{2}} \left(\int_{s^2}^8 e^s dr\right)ds[/MATH]
Is this correct?