5.7 Change of variables in multiple integrals  (Page 7/13)

$x=u^2,y=v^2$

$x=u^3,y=v^3$

$x=u^2,y=v^2,\text{where}S$ is the rectangle of vertices $\left(\mathrm{-1},0\right),\left(1,0\right),\left(1,1\right),\text{and}\left(\mathrm{-1},1\right).$

$x=u^4,y=u^2+v,\text{where}S$ is the triangle of vertices $\left(\mathrm{-2},0\right),\left(2,0\right),\text{and}\left(0,2\right).$

$x=2u,y=3v,\text{where}S$ is the square of vertices $\left(\mathrm{-1},1\right),\left(\mathrm{-1},\mathrm{-1}\right),\left(1,\mathrm{-1}\right),\text{and}\left(1,1\right).$

$T\left(u,v\right)=\left(2u-v,u\right),$ where $S$ is the triangle of vertices $\left(\mathrm{-1},1\right),\left(\mathrm{-1},\mathrm{-1}\right),\text{and}\left(1,\mathrm{-1}\right).$

$x=u+v+w,y=u+v,z=w,$ where $S=R={\text{R}}^3.$

$x=u^2+v+w,y=u^2+v,z=w,$ where $S=R={\text{R}}^3.$

$x=4u,y=5v,$ where $S=R={\text{R}}^2.$

$x=u+2v,y=\text{−}u+v,$ where $S=R={\text{R}}^2.$

$x=e^{2u+v},y=e^{u-v},$ where $S={\text{R}}^2$ and $R=\left\{\left(x,y\right)|x>0,y>0\right\}$

$x=\text{ln}u,y=\text{ln}\left(uv\right),$ where $S=\left\{\left(u,v\right)|u>0,v>0\right\}$ and $R={\text{R}}^2.$

$x=u+v+w,y=3v,z=2w,$ where $S=R={\text{R}}^3.$

$x=u+v,y=v+w,z=u+w,$ where $S=R={\text{R}}^3.$

$x=au,y=bv,R=\left\{\left(x,y\right)|x^2+y^2\le a^2{b}^2\right\},$ where $a,b>0$

$x=au,y=bv,R=\left\{\left(x,y\right)|\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1\right\},$ where $a,b>0$

$x=\frac{u}{a},y=\frac{v}{b},z=\frac{w}{c},$ $R=\left\{\left(x,y\right)|x^2+y^2+z^2\le 1\right\},$ where $a,b,c>0$

$x=au,y=bv,z=cw,R=\left\{\left(x,y\right)|\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}\le 1,z>0\right\},$ where $a,b,c>0$

$x=u+2v,y=\text{−}u+v$

$x=\frac{u^3}{2},y=\frac{v}{u^2}$

$x=e^{2u-v},y=e^{u+v}$

$x=u{e}^v,y=e^{\text{−}v}$

$x=u\text{cos}\left(e^v\right),y=u\text{sin}\left(e^v\right)$

$x=v\text{sin}\left(u^2\right),y=v\text{cos}\left(u^2\right)$

$x=u\text{cosh}v,y=u\text{sinh}v,z=w$

$x=v\text{cosh}\left(\frac{1}{u}\right),y=v\text{sinh}\left(\frac{1}{u}\right),z=u+w^2$

$x=u+v,y=v+w,z=u$

$x=u-v,y=u+v,z=u+v+w$

The triangular region $R$ with the vertices $\left(0,0\right),\left(1,1\right),\text{and}\left(1,2\right)$ is shown in the following figure.

1. Find a transformation $T:S\to R,$ $T\left(u,v\right)=\left(x,y\right)=\left(au+bv,cu+dv\right),$ where $a,b,c,$ and $d$ are real numbers with $ad-bc\ne 0$ such that $T^{\mathrm{-1}}\left(0,0\right)=\left(0,0\right),T^{\mathrm{-1}}\left(1,1\right)=\left(1,0\right),$ and $T^{\mathrm{-1}}\left(1,2\right)=\left(0,1\right).$
2. Use the transformation $T$ to find the area $A\left(R\right)$ of the region $R.$

The triangular region $R$ with the vertices $\left(0,0\right),\left(2,0\right),\text{and}\left(1,3\right)$ is shown in the following figure.

1. Find a transformation $T:S\to R,$ $T\left(u,v\right)=\left(x,y\right)=\left(au+bv,cu+dv\right),$ where $a,b,c$ and $d$ are real numbers with $ad-bc\ne 0$ such that $T^{\mathrm{-1}}\left(0,0\right)=\left(0,0\right),$ $T^{\mathrm{-1}}\left(2,0\right)=\left(1,0\right),$ and $T^{\mathrm{-1}}\left(1,3\right)=\left(0,1\right).$
2. Use the transformation $T$ to find the area $A\left(R\right)$ of the region $R.$

${\displaystyle \underset{R}{\iint }\left(y-x\right)}dA$

${\displaystyle \underset{R}{\iint }\left(y^2-xy\right)}dA$