Integration By Substitution In Higher Dimensions

Applying integration by substitution only works over single integrals. To do it in higher dimensions, we need to involve an extra component: the Jacobian.

Definition §

Integration Substitution (Bivariate) ^definition

Let $f(x,y)$ be a bivariate scalar function, $f:D\subseteq {\mathbb{R}}^2\to \mathbb{R}$. Let $\vec{M}$ be a $C^1$ mapping that transforms coordinate systems, i.e. $T:{\mathbb{R}}^n\to {\mathbb{R}}^n$ but it represents $x,y$ with a different coordinate system. Let the new variables be $(\alpha ,\beta )$. Then $\vec{M}(\alpha ,\beta )=(u(\alpha ,\beta ),v(\alpha ,\beta ))$. , i.e. $x=u(\alpha ,\beta )$ and $y=v(\alpha ,\beta )$.

Then, for a double integral we have:

$${\iint }_{D}f(x,y)\text{d}x\text{d}y={\iint }_{D^{\ast }}f(\vec{M}(\alpha ,\beta ))\cdot \det ({\mathbf{J}}_{\vec{T}})\text{d}\alpha \text{d}\beta$$

• $D^{\ast }$ is the pre-image of $D$ under $\vec{T}$: $\vec{T}:D^{\ast }\to D$
• ${\mathbf{J}}_{\vec{T}}$ is the Jacobi matrix. $\det ({\mathbf{J}}_{\vec{T}})$ is the Jacobian.
• The most common substitution is switching to polar coordinates $(r,\theta )$

Integration Substitution (Trivariate) ^definition

Let $f(x,y,z)$ be a trivariate scalar function, $f:D\subseteq {\mathbb{R}}^3\to \mathbb{R}$. Let $\vec{M}$ be a $C^1$ mapping that transforms coordinate systems, i.e. $T:{\mathbb{R}}^n\to {\mathbb{R}}^n$ but it represents $x,y,z$ with a different coordinate system. Let the new variables be $(\alpha ,\beta ,\gamma )$. Then $\vec{M}(\alpha ,\beta ,\gamma )=(u(\alpha ,\beta ,\gamma ),v(\alpha ,\beta ,\gamma ),w(\alpha ,\beta ,\gamma ))$. , i.e. $x=u(\alpha ,\beta ,\gamma ),y=v(\alpha ,\beta ,\gamma )$ and $z=w(\alpha ,\beta ,\gamma )$

Then, for a triple integral we have:

$${\iiint }_{D}f(x,y,z)\text{d}x\text{d}y\text{d}z={\iiint }_{D^{\ast }}f(\vec{M}(\alpha ,\beta ,\gamma ))\cdot \det ({\mathbf{J}}_{\vec{M}})\text{d}\alpha \text{d}\beta \text{d}\gamma$$

• $D^{\ast }$ is the pre-image of $D$ under $\vec{T}$: $\vec{T}:D^{\ast }\to D$
• ${\mathbf{J}}_{\vec{T}}$ is the Jacobi matrix. $\det ({\mathbf{J}}_{\vec{T}})$ is the Jacobian.
• The most common substitutions are cylindrical coordinates ($\rho ,\varphi ,z$) and spherical coordinates ($r,\theta ,\varphi$)

Integration Substitution (Multivariate) ^definition

Let $f(x_1,x_2,\dots ,x_n)$ be a multivariate scalar function, $f:D\subseteq {\mathbb{R}}^n\to \mathbb{R}$. Let $\vec{M}$ be a $C^1$ mapping that transforms coordinate systems, i.e. $T:{\mathbb{R}}^n\to {\mathbb{R}}^n$ but it represents $x_1,x_2,\dots ,x_n$ with a different coordinate system, $\vec{M}(x_1,\dots ,x_n)=(g_1(y_1,\dots ,y_n),\dots ,g_n(y_1,\dots ,y_n))$

Then, for a given integral (single,double,triple, etc.) we have:

$$\underset{n}{\underbrace{\int \cdots {\int }_D}}f(x_1,\dots ,x_n)\text{d}{x}_1\dots \text{d}{x}_n=\underset{n}{\underbrace{\int \cdots {\int }_{D^{\ast }}}}f(\vec{M}(x_1,\dots ,x_n))\text{d}{x}_1\dots \text{d}{x}_n$$

• $D^{\ast }$ is the pre-image of $D$ under $\vec{T}$: $\vec{T}:D^{\ast }\to D$ #TODO

Common Substitutions §

• Polar Coordinates
• Cylindrical Coordinates
• Spherical Coordinates

Polar Substitution §

Cylindrical Substitution §

Spherical Substitution §

Derivation §

Let’s start by assuming we have a function $f(x,y)$ in Cartesian coordinates, and we need to compute its double integral over some domain $D$. To make things simple, let’s assume that $f:D\to \mathbb{R}$, i.e. $f$ is only defined in $D$.

$${\iint }_{D}f(x,y)\text{d}A={\iint }_{D}f(x,y)\text{d}x\text{d}y$$

And let’s say we want to switch this to a different coordinate system $(\alpha ,\beta )$ instead of $(x,y)$, represented by the conversion:

$$x=u(\alpha ,\beta )y=v(\alpha ,\beta )$$

We can represent this as a mapping (which is just a function, really). Let $\vec{M}(\alpha ,\beta )=(u(\alpha ,\beta ),v(\alpha ,\beta ))$.

Applying $f$ to this new mapping essentially results in a functional composition i.e. we have $f\circ \vec{M}$. The rules of functional composition state that we need the range of $u$ and $v$ to be the domain of $f$ (which is $D$).

So we need $\vec{M}:D^{\ast }\to D$, that is, $\vec{M}$ has some domain $D^{\ast }$ which must map to $D$. $D^{\ast }$ is known as the pre-image of $D$

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This requires $\vec{M}$ to be an injective mapping, since we don’t want two different pairs $(x_1,y_1)$ and $(x_2,y_2)$ to both map to the same $(\alpha ,\beta )$. We also require $\vec{M}$ to be a $C^1$ smooth function (#tosee why?)

Now, consider a small section $\text{d}{A}^{\ast }$ of $D^{\ast }$. We take it to be a rectangle with bottom left corner (${\alpha }_0,{\beta }_0$) and side lengths $\delta \alpha$ and $\delta \beta$.

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What happens to this rectangle after the mapping $\vec{M}$? Under the mapping $\vec{M}$, the bottom and left edges now become parametric curves:

• The bottom left corner now maps to $({\alpha }_0,{\beta }_0)\mapsto (u({\alpha }_0,{\beta }_0),v({\alpha }_0,{\beta }_0))=(x_0,y_0)$
• The bottom edge now becomes a parametric curve $\vec{M}(\alpha ,{\beta }_0)$. Because the original edge had $\beta$ being constant (to the value ${\beta }_0$), the transformed curve also has $\beta$ constant. Hence it is only a single-parameter equation (and hence a curve, and not a parametric surface!).
• Similarly, the left edge is now a parametric edge $\vec{M}({\alpha }_0,\beta )$.

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Hence, we can construct tangent vectors to these parametric curves. Recall the formula for a tangent vector:

So, we have two tangent vectors, ${\vec{T}}_{\alpha }$ and ${\vec{T}}_{\beta }$, at the bottom left corner ${\alpha }_0,{\beta }_0$

$${\vec{T}}_{\alpha }={\left(\frac{\partial u}{\partial \alpha },\frac{\partial v}{\partial \alpha }\right)|}_{({\alpha }_0,{\beta }_0)}{\vec{T}}_{\beta }={\left(\frac{\partial u}{\partial \beta },\frac{\partial v}{\partial \beta }\right)|}_{({\alpha }_0,{\beta }_0)}$$

Note that we are now in the Cartesian coordinate space, and we are treating $\alpha$ and $\beta$ as parameters!

We can approximate the area of the transformed rectangle by using the cross product of the tangent vectors, scaled by $\delta$. This is because the geometry of cross product produces a parallelogram, and the magnitude is equivalent to the area of the paralellogram.

$$\Vert \delta \alpha \cdot {\vec{T}}_{\alpha }\times \delta \beta \cdot {\vec{T}}_{\beta }\Vert =\delta \alpha \cdot \delta \beta \cdot \Vert {\vec{T}}_{\alpha }\times {\vec{T}}_{\beta }\Vert$$

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The cross product expands to

$$\begin{array}{rl}\delta \alpha \cdot \delta \beta \cdot \Vert {\vec{T}}_{\alpha }\times {\vec{T}}_{\beta }\Vert & =\delta \alpha \cdot \delta \beta \cdot \Vert \det \left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial u}{\partial \alpha }& \frac{\partial v}{\partial \alpha }& 0\\ \frac{\partial u}{\partial \beta }& \frac{\partial v}{\partial \beta }& 0\end{array}\right]\Vert \\ & =\delta \alpha \cdot \delta \beta \cdot \Vert (0\hat{i}-0\hat{j}+\det \left[\begin{array}{cc}\frac{\partial u}{\partial \alpha }& \frac{\partial v}{\partial \alpha }\\ \frac{\partial u}{\partial \beta }& \frac{\partial v}{\partial \beta }\end{array}\right]\hat{k})\Vert \\ & =\delta \alpha \cdot \delta \beta \cdot \det \left[\begin{array}{cc}\frac{\partial u}{\partial \alpha }& \frac{\partial v}{\partial \alpha }\\ \frac{\partial u}{\partial \beta }& \frac{\partial v}{\partial \beta }\end{array}\right]\\ & =\delta \alpha \cdot \delta \beta \cdot \det \left[\begin{array}{cc}\frac{\partial u}{\partial \alpha }& \frac{\partial u}{\partial \beta }\\ \frac{\partial v}{\partial \alpha }& \frac{\partial v}{\partial \beta }\end{array}\right]\\ & =\delta \alpha \cdot \delta \beta \cdot \det ({\mathbf{J}}_{\vec{M}})\end{array}\text{\hspace{1em}(We can swap diagonals)}$$

The Jacobian reappears!

When we integrate, this approximation becomes closer and closer to the actual $\text{d}A$, until they are equal. Then we have:

$$\text{d}A=\text{d}x\text{d}y=\det ({\mathbf{J}}_{\vec{M}})\text{d}\alpha \text{d}\beta$$

So we finally have:

$${\iint }_{D}f(x,y)\text{d}A={\iint }_{D^{\ast }}f(\vec{M}(\alpha ,\beta ))\cdot \det ({\mathbf{J}}_{\vec{M}})\text{d}\alpha \text{d}\beta$$