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# Change of variables in integral

## 1 Indefinite integral

techniques of integration...

## 2 Definite integral

Consider how we compute this integral in two ways:

$$\displaystyle\int_0^1 2x e^{x^2} dx.$$

In either case we use change of variables.

(1) We use the substitution:

$$u = x^2, du = 2xdx.$$

to compute the antiderivative (with respect to $x$):

$$\displaystyle\int 2x e^{x^2} dx= \displaystyle\int e^u du = e^u + c = e^{x^2} + C,$$

and then apply the Fundamental Theorem of Calculus with respect to $x$:

$$\displaystyle\int_0^1 2x e^{x^2} dx = e^{x^2} |_0^1 = e - 1.$$

(2) Alternatively, we use the same substitution to get

$$\displaystyle\int_{x=0}^{x=1} 2x e^{x^2} dx = \displaystyle\int_{u=0}^{u=1} e^u du$$

but don't do "the back-substitution" and then apply the Fundamental Theorem of Calculus with respect to $u$:

$$= \displaystyle\int_0^1 e^u du$$ $$= e^u |_0^1$$ $$= e^1 - e^0.$$

## 3 Definite integral in polar coordinates

We want to utilize the geometry of polar coordinate system to make integration over some (curved) regions simpler.

$$a_1 {\leq} x {\leq} b_1, a_2 {\leq} y {\leq} b_2.$$

Now, we follow the same procedure with polar coordinates:

$$r_0 {\leq} r {\leq} r_1, {\theta}_0 {\leq} {\theta} {\leq} {\theta}_1$$

The geometry of this "rectangle" is different of course. In fact, we have

a circle: $0 {\leq} r {\leq} 1, 0 {\leq} {\theta} {\leq} 2{\pi}$.

To compute

$$\displaystyle\int\displaystyle\int_G f( x, y ) dA$$

in polar coordinates (volume under the graph) we need some kind of Riemann sum. What we do exactly mimics the rectangular case, but only algebraically:

• divide $[ {\theta}_0, {\theta}_1 ]$ into $k$ intervals,
• length ${\Delta}{\theta}$,
• divide $[ r_0, r_1 ]$ into $k$ intervals,
• length ${\Delta}r$,
• evaluate $f( r_i, {\theta}_j )$ with each $( r_i, {\theta}_j )$ in $R_{ij}$.

Then the Riemann sum is

$$\displaystyle\sum_{i=1}^k \displaystyle\sum_{j=1}^k f( r_i, {\theta}_j ) {\Delta}A,$$

where ${\Delta}A$ is the area of the rectangle $R_{ij}$.

Approximate the volume by the sum of these volumes,

volume $=$ height $\cdot$ area of the base

We compute the area of the curved rectangle with sides ${\Delta}r, {\Delta}{\theta}$, which varies with the location (unlike rectangular). How? Consider a circle with radius $r$ and area ${\pi}r^2$, and another circle with radius $r - {\Delta}r$ and area ${\pi}( r - {\Delta}r )^2$. Then we obtain a ring with area

$$\begin{array}{} {\pi}r^2 - {\pi}( r - {\Delta}r )^2 &= {\pi}( r^2 - ( r - {\Delta}r )^2 ) \\ &= {\pi}( 2r - {\Delta}r ) {\Delta}r \\ &= ( r - \frac{{\Delta}r}{2} ) {\Delta}r {\Delta}{\theta}. \end{array}$$

"Rectangle", $R$, the area is equal the the area of a ring multiplied by $\frac{{\Delta}{\theta}}{2{\pi}}$, hence

area $= \frac{1}{2}( 2r - {\Delta}r ) {\Delta}r {\Delta}{\theta}, {\theta}_i {\leq} {\theta} {\leq} {\theta}_i + {\Delta}{\theta}.$

The total volume is approximated by the Riemann sum

$$\begin{array}{} \displaystyle\sum_{i,j}^k f( r_i, {\theta}_j ) ( r - \frac{{\Delta}r}{2} ) {\Delta}r {\Delta}{\theta} &= ... \\ &= \lim_{k \rightarrow \infty} \displaystyle\sum_{i,j}^k f( r_i, {\theta}_j ) ( r - \frac{{\Delta}r}{2} ) {\Delta}r {\Delta}{\theta} \\ &= \displaystyle\int_{\theta_0}^{\theta_1} \displaystyle\int_{r_0}^{r_1} f( r, {\theta} ) r dr d{\theta}. \end{array}$$

Example. Compute the volume of the half-sphere

$$\displaystyle\int\displaystyle\int_D ( 1 - x^2 - y^2 )^{\frac{1}{2}} dA$$

with $D: 0 {\leq} {\theta} {\leq} 2{\pi}, 0 {\leq} r {\leq} 1$. Further

$$g( x, y ) {\rightarrow} f( r, {\theta} ),$$

$$z = ( 1 - x^2 - y^2 )^{\frac{1}{2}} {\rightarrow} ( 1 - ( r \cos {\theta} )^2 - ( \sin {\theta} )^2 )^{\frac{1}{2}} = ( 1 - r^2 )^{\frac{1}{2}}.$$

Then

$$\begin{array}{} V &= \displaystyle\int_{\theta=0}^{\theta = 2 \pi} \displaystyle\int_{r=0}^{r=1} ( 1 - r^2 )^{\frac{1}{2}} r d{\theta} dr &= 2{\pi} \displaystyle\int_0^1 ( 1 - r^2 )^{\frac{1}{2}} r dr. \end{array}$$

Finally, with

$$u = 1 - r^2 , du = -2 r dr$$\

we have

$$\begin{array}{} 2{\pi} \displaystyle\int_0^1 u^{\frac{1}{2}} ( -1 / 2 ) du &= \frac{2{\pi}}{2} \cdot \frac{2}{3} \cdot u^{\frac{3}{2}} |_0^1 \\ &= \frac{2}{3} \cdot {\pi}. \end{array}$$

Example. Compute the area of a circle:

$A =$ the volume under the graph of $f( x, y ) = 1$.

$$\begin{array}{} A = V &= \displaystyle\int\displaystyle\int_D 1 dA \\ &= \displaystyle\int_0^{2 \pi} \displaystyle\int_0^1 r dr d{\theta} \\ &= \displaystyle\int_0^{2 \pi} \left. \frac{r^2}{2} \right|_0^1 d{\theta} \\ &= \displaystyle\int_0^{2 \pi} \frac{1}{2} d{\theta} \\ &= 2{\pi} \cdot \frac{1}{2} \\ &= {\pi}. \end{array}$$

Note that in

$$\displaystyle\int_Q g( x, y ) dA = \displaystyle\int_R g( r \cos {\theta}, r \sin {\theta} ) r dr d{\theta},$$

$g( r \cos {\theta}, r \sin {\theta} ) r$ is a composition.

## 4 Definite integral in dimension $n$

Let ${\rm \hspace{3pt} dim \hspace{3pt}} = 2$. Compute the integral

$$\displaystyle\int_B f( x, y ) dA$$

with respect to another coordinate system $( u, v )$. This system comes from a transformation

$$( x, y ) = T( u, v )$$

with $T: {\bf R}^2 {\rightarrow} {\bf R}^2$.

The question is: what is the function (of $u,v$) we should integrate over $Q$ to get the same result as integration of $f(x,y)$ over $B = T(Q)$:

$$\displaystyle\int_Q ... dA = \displaystyle\int_B f( x, y ) dA?$$

Compare the Riemann sums:

$$\displaystyle\sum_i f( T( u_i, v_i ) ) {\Delta}W =^{!!} \displaystyle\sum_i f( x_i, y_i ) {\Delta}A,$$

where the $u_i$ are the heights, ${\Delta}W$ the areas, the $x_i$ the heights and ${\Delta}A$ the area of the base. Then what is the correspondence:

$$T( u_i, v_i ) {\longleftrightarrow}^? ( x_i, y_i )$$

This is what we do know:

$$9 {\Delta}W {\longleftrightarrow} {\Delta}A$$

Then

$$\displaystyle\sum_i f( T( u_i, v_i ) ) 9 {\Delta}W = \displaystyle\sum_i f( x_i, y_i ) {\Delta}A,$$

where $( 9 {\Delta}W )$ equals ${\Delta}A$. Substitution and the limit and the Riemann sum of $f$, respectively, yield

$$\displaystyle\int_Q f( T( u, v )) 9 dW = \displaystyle\int_B f( x, y ) dA.$$

Where does this 9 come from?

$$T( u, v ) = \left| \begin{array}{} 3 & 0 \\ 0 & 3 \end{array} \right| \left| \begin{array}{} 4 \\ v \end{array} \right|$$

and

$$T' = \left| \begin{array}{} 3 & 0 \\ 0 & 3 \end{array} \right|$$

Then $9 = | {\rm det \hspace{3pt}} T' |$. This is the "stretching coefficient".

Let's consider $| {\rm det \hspace{3pt}} T' |$ in a couple of examples.

Example. Polar coordinates:

$$\displaystyle\int_B f( x, y ) dA = \displaystyle\int\displaystyle\int f( r \cos {\theta}, r \sin {\theta} ) r dr d{\theta},$$

$$T( r, {\theta} ) = ( r \cos {\theta}, r \sin {\theta} )$$

$$T'( r, {\theta} ) = \left| \begin{array}{} \cos \theta & -\sin \theta \\ \sin \theta & r \cos \theta \end{array} \right|$$

and

$$\begin{array}{} | {\rm det \hspace{3pt}} T'( r, {\theta} ) | &= | \cos {\theta} r \cos {\theta} - \sin {\theta} ( -r \sin {\theta} ) | \\ &= | r \cos^2 {\theta} + r \sin^2 {\theta} | \\ &= r | \cos^2 {\theta} + \sin^2 {\theta} | \\ &= r. \end{array}$$

Example. Rotation, rotated through ${\alpha}$

$$T' = \left| \begin{array}{} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array} \right|$$

$${\rm det \hspace{3pt}} T' = \cos^2 {\alpha} + \sin^2 {\alpha} = 1,$$

no stretching.

Review exercise. Consider

$$\displaystyle\int\displaystyle\int_{B(0,1)} x^2 y^2 dA,$$

convert to iterated integral:

$$\displaystyle\int_{-1}^1 \displaystyle\int_{\sqrt{1-x^2}}^{\sqrt{1-x^2}} x^2 y^2 dy dx$$

Consider a more general situation. Suppose

$T: {\bf R}^2 {\rightarrow} {\bf R}^2$ is linear.

Then

area $B = || v_1 \times v_2 || \longleftrightarrow v_1 = \left| \begin{array}{} * \\ * \\ 0 \end{array} \right|, v_2 = \left| \begin{array}{} * \\ * \\ 0 \end{array} \right|$

Further

$$v_1 = T( e_1 ), v_2 T( e_2 ).$$

Now

$$T(u,v) = \left| \begin{array}{} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| \left| \begin{array}{} u \\ v \end{array} \right|$$

then

$$v_1 = T(e_1) = \left| \begin{array}{} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| \left| \begin{array}{} 1 \\ 0 \end{array} \right| = \left| \begin{array}{} a_{11} \\ a_{21} \end{array} \right|$$

and

$$v_1 = T(e_2) = \left| \begin{array}{} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| \left| \begin{array}{} 0 \\ 1 \end{array} \right| = \left| \begin{array}{} a_{12} \\ a_{22} \end{array} \right|$$

Substitute:

$$\begin{array}{} {\rm area \hspace{3pt}} B = ||v_1 \times v_2|| &= \left\lVert \begin{array}{ccc} |a_{11}| &\times &|a_{12}| \\ |a_{21}| &\times &|a_{22}| \\ |0| &\times &|0| \end{array} \right\rVert \\ &= \left\lVert {\rm det} \left| \begin{array}{} a_{11} & a_{12} & i \\ a_{21} & a_{22} & j \\ 0 & 0 & k \end{array} \right| \right\rVert \\ &=\lVert 0i + 0j + {\rm det \hspace{3pt}} M_t k \rVert \\ &=| \det M_t|, \end{array}$$

where $M_t$ is the matrix of $T$.

Theorem.

<center>area $T(Q) = | \det( M_t ) |$

where $T$ is affine, $M_t$ is its matrix, $Q$ the unit square.

Riemann sum of $f$ equals

$$\displaystyle\sum_i f( x_i, y_i ) {\Delta}A,$$

where ${\Delta}A$ are the areas of the parallelograms.

Transform it to the Riemann sum over $( u, v )$

$$\displaystyle\sum_i f( T( u, v_i ) ) | \det T' | {\Delta}W,$$

where ${\Delta}W$ is the area of the squares.

Theorem. If $T$ is an affine function $T: {\bf R}^2 {\rightarrow} {\bf R}^2$, and $Q$ a square, $B = T(Q)$, then

$$\displaystyle\int\displaystyle\int_{B=T(Q)} f( x, y ) dA = \displaystyle\int\displaystyle\int_Q f( T( u, v ) ) | \det T' | dW.$$

Problem. Suppose $B$ is the parallelogram on the right. Evaluate $\displaystyle\int\displaystyle\int_B x^2 y dA$.

We could solve the problem by applying iterated integral with these bounds as graphs bounding $B$ from below and above. But instead we will simplify $B$, then integrate over $Q$.

How do we transform $B$ into the unit square? We need to find $S: {\bf R}^2 {\rightarrow} {\bf R}^2$ such that

$$S(B) = Q.$$

Turns out $T = S^{-1}$ if

$$T(u,v) = \left| \begin{array}{} 1 & 3 \\ 2 & 1 \end{array} \right| \left| \begin{array}{} u \\ v \end{array} \right| + \left| \begin{array}{} 1 \\ 1 \end{array} \right|.$$

To verify that $T$ is an invertible function, check:

$$\det T \neq 0,$$

Indeed:

$$| \det T | = \left| \begin{array} 1 & 3 \\ 2 & 1 \end{array} \right| = | 1 - 6| = 5.$$

We have $f( x, y ) = x^2 y$ and

$$\displaystyle\int\displaystyle\int_B x^2 y dA = \displaystyle\int\displaystyle\int_Q f T( u, v ) 5 dW {\rm \hspace{3pt} (iterated)}$$

Then $$\begin{array}{} f T(u,v) &= f \left( \left| \begin{array}{} 1 & 3 \\ 2 & 1 \end{array} \right| \left| \begin{array}{} u \\ v \end{array} \right| + \left| \begin{array}{} 1 \\ 1 \end{array} \right| \right) \\ &= f(u + 3v + 1, 2u+v+1 ) \\ &= (u + 3v + 1)^2 (2u+v+1), \end{array}$$

hence

$$\displaystyle\int\displaystyle\int_B x^2 y dA = \displaystyle\int_0^1 \displaystyle\int_0^1 ( u + 3v + 1 )^2 ( 2u + v + 1 ) du dv.$$

The rest of the calculation is routine.

Exercise. A similar example for integration over a disk:

$$\displaystyle\int\displaystyle\int_{B(0,1)} x^2 y^2 dA,$$

with

$$x = r \cos {\theta}$$

$$y = r \sin {\theta}$$

and $| det T' | = r$. Further

$$\begin{array}{} \displaystyle\int_0^{2 \pi} \displaystyle\int_0^1 (r \cos {\theta} )^2 (r \sin {\theta} )^2 r dr d{\theta} &= \displaystyle\int_0^{2 \pi} \displaystyle\int_0^1 r^5 \cos {\theta} \sin {\theta} dr d{\theta} \\ &= \displaystyle\int_0^{2 \pi} \cos^2 {\theta} \sin^2 {\theta} d{\theta} \displaystyle\int_0^1 r^5 dr \\ &= ... \end{array}$$

substitute $u = \sin {\theta}$ and complete the computation.

What if $F$ is not affine?

Then find the best affine approximation $T$ of $F$, at each point $( a, b )$. Which means that

$$F'( a, b ) = T'( a, b ).$$

This approximation applies to every element of the Riemann sum, i.e., on each of the rectangles $R_{ij}$, in total

$\displaystyle\sum_{i,j} \displaystyle\int\displaystyle\int_{R_{ij}} f( T( u, v )) | \det T'( a, b )| dW = \displaystyle\int\displaystyle\int_{R_{ij}} f( T( u, v )) | \det T'( u, v )| dW$ (using the limit and the additivity property)

And the result is:

$$\displaystyle\int\displaystyle\int_{B=T(Q)} f( x, y ) dA = \displaystyle\int\displaystyle\int_Q f( T( u, v )) | \det' T( u, v )| dW.$$