Substitutions in Multiple Integrals

Substitutions in double integrals

Recall the one-dimensional case $$\int_{g(a)}^{g(b)}f(x)dx=\int_a^b f(g(u))g'(u)du$$

Suppose that a region $G$ in the $uv$-plane is transformed into the region $R$ in the $xy$-plane by $$x=g(u,v),\quad y=h(u,v).$$

We assume the transformation is one-to-one on the interior of $G$.

• image of $G$: $R$

• preimage of $R$: $G$ .

Jacobian

definition

The Jacobian determinant or Jacobian of the coordinate transformation $x = g(u, v), y = h(u, v)$ is $$J(u,v)={\partial (x,y)\over \partial (u,v)}=\begin{vmatrix} {\partial x\over\partial u} & {\partial x\over\partial v}\\ {\partial y\over\partial u} & {\partial y\over\partial v} \end{vmatrix}={\partial x\over\partial u}{\partial y\over\partial v}-{\partial y\over\partial u}{\partial x\over\partial v}.$$

Theorem

Suppose that $f(x, y)$ is continuous over the region $R$. Let $G$ be the preimage of R under the transformation $x = g(u, v), y = h(u, v)$, assumed to be one-to-one on the interior of $G$. If the functions $g$ and $h$ have continuous first partial derivatives within the interior of $G$, then $$\iint_Rf(x,y)dxdy=\iint_G f(g(u,v),h(u,v))\left|{\partial (x,y)\over\partial (u,v)}\right|dudv.$$

Example1: evaluate $$\int_0^4\int_{x=y/2}^{x=(y/2)+1}{2x-y\over 2}dxdy$$ by the transformation $$u={2x-y\over 2}, \quad v={y\over 2}.$$ Solution:

From the transformation equations, we know that: $$x=u+v, y=2v$$ $xy$-equations for the boundary of $R \rightarrow$ $uv$-equations for the boundary of $G$:

• $x=y/2 \rightarrow u=0$

• $x=y/2+1 \rightarrow u=1$

• $y=0 \rightarrow v=0$

• $y=4 \rightarrow v=2$

Example2: evaluate $$\int_0^1\int_0^{1-x}{\sqrt{x+y}(y-2x)^2}dydx$$ Solution:

Here we need to define the transformation equations by ourselves. We want to simplify the integral formula while making the boundary of G easy to find. Thus we can define $$u=x+y,v=y-2x$$ And from the transformation equations, we know that: $$x={{u-v}\over 3}, y={2u+v\over 3}$$ $xy$-equations for the boundary of $R \rightarrow$ $uv$-equations for the boundary of $G$:

• $x=0 \rightarrow v=u$

• $x=1 \rightarrow v=u-3$

• $y=0 \rightarrow v=-2u$

• $y=1-x \rightarrow u=1$

Example3: evaluate $$\int_1^2\int_{1/y}^y\sqrt{y\over x}e^{\sqrt{xy}}dxdy$$ Solution:

We can define $$u=\sqrt{y\over x},v=\sqrt{xy}$$ And from the transformation equations, we know that: $$x={v\over u}, y=uv$$ $xy$-equations for the boundary of $R \rightarrow$ $uv$-equations for the boundary of $G$:

• $y\geq 1 \rightarrow uv\geq 1$

• $y\leq 2 \rightarrow uv\leq 2$

• $x\geq 1/y \rightarrow v\geq 1$

• $x\leq y \rightarrow u\geq 1$

Substitutions in triple integrals

Cylindrical coordinates: $$x=r\cos\theta,~y=r\sin\theta,~z=z.$$ $$J(r,\theta,z)={\partial (x,y,z)\over \partial (r,\theta,z)}=\begin{vmatrix} {\partial x\over\partial r} & {\partial x\over\partial \theta} & {\partial x\over\partial z}\\ {\partial y\over\partial r} & {\partial y\over\partial \theta} & {\partial y\over\partial z}\\ {\partial z\over\partial r} & {\partial z\over\partial \theta} & {\partial z\over\partial z} \end{vmatrix} =\begin{vmatrix} {\cos \theta} & {-r \sin \theta} & {0}\\ {\sin \theta} & {r \cos \theta} & {0}\\ {0} & {0} & {1} \end{vmatrix} =r$$ Spherical coordinates:$$x=\rho\sin\phi\cos\theta,~y=\rho\sin\phi\sin\theta,~z=\rho\cos\phi.$$ $$J(\rho,\phi,\theta)={\partial (x,y,z)\over \partial (\rho,\phi,\theta)}=\begin{vmatrix} {\partial x\over\partial \rho} & {\partial x\over\partial \phi} & {\partial x\over\partial \theta}\\ {\partial y\over\partial \rho} & {\partial y\over\partial \phi} & {\partial y\over\partial \theta}\\ {\partial z\over\partial \rho} & {\partial z\over\partial \phi} & {\partial z\over\partial \theta} \end{vmatrix} =\begin{vmatrix} {\sin \phi \cos \theta} & {\rho \cos \phi \cos \theta} & {- \rho \sin \phi \sin \theta}\\ {\sin \phi \sin \theta} & {\rho \cos \phi \sin \theta} & {\rho \sin \phi \cos \theta}\\ {\cos \phi} & {-\rho \sin \phi} & {0} \end{vmatrix} =\rho^2 \sin \phi$$

Example: Evaluate $$\int_0^3\int_0^4\int_{x=y/2}^{x=(y/2)+1}\left({2x-y\over2}+{z\over3}\right)dxdydz$$ with transformation $$u=(2x-y)/2,~v=y/2,~w=z/3.$$ From the transformation equations, we know that: $$x=u+v, y=2v,z=3w$$ $xyz$-equations for the boundary of $R \rightarrow$ $uvw$-equations for the boundary of $G$:

• $x=y/2 \rightarrow u=0$

• $x=y/2+1 \rightarrow u=1$

• $y=0 \rightarrow v=0$

• $y=4 \rightarrow v=2$

• $z=0 \rightarrow w=0$

• $z=3 \rightarrow w=1$