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The things you own end up owning you, Fight Club
A double integral allows us to integrate a function of two variables, say f(x, y), over a region R in a two-dimensional space (e.g. in the xy-plane). Instead of just finding or calculating an area, with a double integral, we’re looking for the volume under the surface (or below the graph) $z = f(x, y)$. It is denoted or written as $\iint_R f(x, y) dA$
Fubini’s Theorem. Suppose $f(x, y)$ is a continuous function defined on a rectangular region $R = [a, b] \times [c, d]$ in the xy-plane. Fubini’s Theorem states that the double integral of $f(x, y)$ over the region R can be computed as an iterated integral in either of two orders: $\int \int_R f(x, y)dA = \int_{a}^{b}\int_{c}^{d} f(x, y)dydx = \int_{c}^{d}\int_{a}^{b} f(x, y)dxdy$
This is a powerful technique, particularly when dealing with double integrals over regions that are challenging to compute using standard Cartesian coordinates. By transforming the coordinates to a different coordinate system, such as polar, cylindrical, or spherical coordinates, or through a custom general transformation, the integral can often be drastically simplified, making an otherwise impossible calculation manageable.
Let’s explore how this technique works by calculating the area of an ellipse. An ellipse with semi-major axis a and semi-minor axis b is described by the equation $(\frac{x}{a})^2 + (\frac{y}{b})^2 \leq 1$. If r = a = b, this equation describes a circle of radius $r$.
However, for a general ellipse, we aim to compute the area by evaluating the double integral: $ \iint_{(\frac{x}{a})^2 + (\frac{y}{b})^2 \leq 1} dxdy$
The previous change of variables was pretty easy, but this is not always the case. When we change variables in a double integral, we must account for how the area element $dA = dxdy$ in the original coordinates transforms into a new area element $dA'=dudv$ in the new coordinates. Space can be stretched, compressed, or rotated. The scaling factor that correctly adjusts the area element for this transformation is provided by the Jacobian determinant.
Consider a linear transformation mapping a region $R$ in the $xy$-plane to a region $S$ in the $uv$-plane, defined by: $u = 3x - 2y$, $v = x + y$.
This transformation maps a region $R$ in the xy-plane to a region $S$ in the uv-plane. To understand how the area element $dA = dxdy$ in the xy-plane relates to the area element $dA′ = dudv$ in the uv-plane, we need to explore how the linear transformation affects the geometry of the region.
Let’s consider a small rectangle $\Delta A$ in the $xy$-plane. Because our mapping is linear, this rectangle is mapped to a parallelogram $\Delta A'$ in the $uv$-plane. 🎯 The critical property of linear transformations is that they scale areas by a constant factor that is independent of the specific location of the rectangle (Figure 1 and 2).
In geometric terms, if we consider two vectors representing the sides of the rectangle $\Delta A$ in the $xy$-plane, their images under the transformation will become the vectors defining the parallelogram $\Delta A'$ in the $uv$-plane.
Recall from linear algebra that the area of a parallelogram spanned by two vectors $\mathbf{u} = \langle u_1, u_2 \rangle$ and $\mathbf{v} = \langle v_1, v_2 \rangle$ is given by the absolute value of the determinant of their matrix: $\text{Area} = \left\vert{} \det \begin{pmatrix} u_1 & v_1 \\ u_2 & v_2 \end{pmatrix} \right\vert{} = \begin{vmatrix} u_1 & v_1 \\ u_2 & v_2 \end{vmatrix}$.
For our transformation, we compute how the new variables change with respect to the old variables: $J_{uv \to xy} = \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \begin{vmatrix} 3 & -2 \\ 1 & 1 \end{vmatrix}$
The area of that parallelogram is $\text{Area}(\Delta A') = \left\vert{} \det \begin{pmatrix} 3 & -2 \\ 1 & 1 \end{pmatrix} \right\vert{} \cdot (\Delta x \Delta y)$. So every tiny rectangle is stretched by a factor of 5, no matter where it is.
The determinant is $(3)(1) - (-2)(1) = 5$. This means the transformation from $(x,y)$ to $(u,v)$ scales areas by a factor of 5. Therefore, $dudv = 5 dxdy$.
However, when evaluating integrals, we substitute $dxdy$ out. Rearranging the relationship gives us $dxdy = \frac{1}{5} \,dudv$.
The change of variables formula for double integrals states that if you have a transformation that maps a region R in the xy-plane to a region S in the uv-plane, then the double integral over R can be computed as an integral over S, with the area element adjusted by the Jacobian determinant: $ \iint_R f(x, y)dA = \iint_S f(x(u,v), y(u, v))|\frac{1}{det(J)}| \cdot dA'$.
For our particular example, where $det(J) = 5$, the integral becomes: $\iint_R f(x, y)dA = \iint_S f(x(u,v), y(u, v))\frac{1}{5}dA' = \frac{1}{5}\iint_S f(x(u,v), y(u, v))dudv$.
In the general case, consider two functions $u = u(x, y), v = v(x, y)$, which define a transformation from the xy-plane to the uv-plane. Small changes $\Delta x$ and $\Delta y$ in the xy-plane lead to corresponding changes $\Delta u$ and $\Delta v$ in the uv-plane, which can be approximated as: $\Delta u \approx \frac{\partial u}{\partial x}\Delta x + \frac{\partial u}{\partial y}\Delta y = u_x\Delta x + u_y\Delta y$ where $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}$ are the partial derivatives of u with respect to x and y respectively (they measure how the function u changes as each variable changes, while holding the other variables constant), and similarly $\Delta v \approx v_x\Delta x + v_y\Delta y$.
This relationship can be written or expressed in matrix form: $\begin{pmatrix}\Delta u\\ \Delta v\end{pmatrix} = \begin{pmatrix}u_x & u_y\\ v_x & v_y\end{pmatrix}\begin{pmatrix}\Delta x\\ \Delta y\end{pmatrix}$ (Figure 4).
Take a tiny rectangle with sides $\Delta x$ and $\Delta y$. This rectangle is spanned by the vectors: $\Delta x e_x = (\Delta x, 0), \Delta y e_y = (0, \Delta y)$. Its area is simply $\Delta A = \Delta x \Delta y$. Under the map $u = u(x, y), v = v(x, y)$, the basis vectors transform according to the partial derivatives: $T(e_x) = (u_x, v_x), T(e_y) = (u_y, v_y)$. These two vectors are exactly the columns of the Jacobian matrix. So the tiny rectangle ($\Delta A'$) becomes a tiny parallelogram spanned by: $\Delta x(u_x, v_x), \Delta y(u_y, v_y)$, and its area is $\Delta A' = \left\vert{}\begin{pmatrix}u_x & u_y\\ v_x & v_y\end{pmatrix} \right\vert{} \cdot (\Delta x \Delta y)$.
The determinant of this Jacobian matrix J gives the scaling factor for the area element: det(J) = $det(\frac{\partial(u,v)}{\partial(x,y)}) = \left\vert{}\begin{pmatrix}u_x & u_y\\ v_x & v_y\end{pmatrix} \right\vert{} \implies dA' = \vert{}J\vert{}dA \implies dudv = \vert{}J\vert{}dxdy= \left\vert{} \begin{pmatrix}u_x & u_y \\ v_x & v_y\end{pmatrix}\right\vert{} dxdy$.
Thus, the differential area elements transform as: $dA' = \vert{}det(J)\vert{}dA$ or $dudv = \vert{}det(J)\vert{}dxdy$.
To sum up, the change of variables formula for double integrals is a powerful tool that allows us to simplify integrals by transforming the region of integration. Given a transformation u = u(x, y) and v = v(x, y) that maps a region R in the xy-plane to a region S in the uv-plane, then the double integral over R is be computed by transforming it into an integral over S as: $\boxed{\iint_R f(x, y)dA = \iint_S f(x(u,v), y(u, v)) \cdot \left\vert{} \frac{1}{det(J)} \right\vert{} dudv = \iint_S f(x(u,v), y(u, v)) \cdot \left\vert{} \frac{1}{det\begin{pmatrix}u_x & u_y\\ v_x & v_y\end{pmatrix}} \right\vert{} dudv}$.
Here, $f(x, y)$ represents the function to be integrated over R, $f(x(u, v), y(u, v))$ is the same function expressed in terms of the new variables u and v and $|\frac{1}{det(J)}| = \frac{1}{\left\vert{} det\begin{pmatrix}u_x & u_y\\ v_x & v_y\end{pmatrix} \right\vert{}}$ is the reciprocal of the determinant of the Jacobian matrix of the transformation. It is a factor that adjusts the area element for the change in the scale caused by the transformation.
If you need to reverse the transformation, the product of the Jacobian determinant of the original transformation and its inverse is always equal to 1: $\frac{\partial(u,v)}{\partial(x,y)} \cdot \frac{\partial(x,y)}{\partial(u,v)} = 1$
Polar coordinates are particularly useful when dealing with regions that have circular symmetry, such as circles or sectors of circles. The process of transforming an integral from Cartesian coordinates $(x,y)$ to polar coordinates $(r,\theta)$ requires adjusting the differential area element appropriately, which is where the Jacobian determinant comes into play. It provides the necessary factor to correctly adjust the differential area element.
The relationship between Cartesian coordinates $(x, y)$ and polar coordinates $(r, \theta)$ is given by the following equations:$x = r\cos(\theta), y = r\sin(\theta)$ where $r$ represents the radial distance from the origin, and $\theta$ is the angle measured counterclockwise from the positive x-axis.
To transform the differential area element $dA = dxdy$ in Cartesian coordinates to polar coordinates, we need to calculate the Jacobian determinant of the transformation.
The Jacobian matrix of this transformation is constructed by taking the partial derivatives of x and y with respect to r and $\theta$ respectively: $J = \frac{\partial(x, y)}{\partial(r, \theta)} = \begin{pmatrix}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{pmatrix} = \begin{pmatrix}x_r & x_\theta \\ y_r & y_\theta \end{pmatrix} = \begin{pmatrix}\cos(\theta) & -r\sin(\theta)\\ \sin(\theta) & r\cos(\theta)\end{pmatrix}.$
Evaluating the determinant: $det(J) = r\cos^2(\theta)+r\sin^2(\theta) = r(\cos^2(\theta)+\sin^2(\theta)) = r·1 = r \implies dxdy = \vert{} r \vert{} drd\theta$ =[r is always positive] $rdrd\theta$. The Jacobian determinant for this transformation is r, so the area element dA in polar coordinates becomes $rdrd\theta$ ($dA = rdrd\theta$). Conversely, if you start in polar coordinates and wish to convert an integral back to Cartesian coordinates, you use the inverse of the Jacobian determinant, which for this transformation would simply be $\frac{1}{r}$.
Identify the new coordinates: $u = x, v = xy$
Determine the Jacobian determinant, $\frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ y & x \end{vmatrix} = (1)(x) - (0)(y) = x$. This determinant is crucial as it gives us the scaling factor for the area element in the new coordinates.
Using the inverse property, the scaling factor we need is $\frac{1}{x}$. Thus, $dxdy = \frac{1}{x} \,dudv$.
Determine the New Limits of Integration.
This is often the trickiest step. The original region $R$ in the $xy$-plane is a unit square bounded by $x=0$, $x=1$, $y=0$, and $y=1$. Let’s map these four boundaries to the $uv$-plane:
$\star$ Line $x=0$: $u=0$, and $v=0(y)=0$. (Maps to the origin).
$\star$ Line $x=1$: $u=1$, and $v=1(y)=y$. Since $y \in [0,1]$, this maps to the vertical line segment $u=1$ for $v \in [0,1]$.
$\star$ Line $y=0$: $v=x(0)=0$, and $u=x$. Since $x \in [0,1]$, this maps to the horizontal line segment $v=0$ for $u \in [0,1]$.
$\star$ Line $y=1$: $v=x(1)=x$. Since $u=x$, this boundary maps to the diagonal line $v=u$.
Thus, our new region $S$ in the $uv$-plane is a triangle bounded by $v=0$, $u=1$, and $v=u$.
If we integrate with respect to $u$ first (horizontal slices), $v$ ranges from $0$ to $1$. For any fixed $v$, $u$ starts at the diagonal line $u=v$ and ends at the vertical line $u=1$.
Set up the integral in the new coordinates. The differential area $dA = dxdy = \frac{1}{x}dudv$. The integrand in terms of u, v: $x^2ydxdy = x^2y\frac{1}{x}dudv = xydudv = vdudv$. Therefore, the integral transforms to: $\int_{0}^{1}\int_{0}^{1} x^2ydxdy = \int_{0}^{1}\int_{v}^{1} vdudv$ (Figure 1, Figure 2 illustrates another way of looking at the same problem).
Integrate. The inner integral is $\int_{v}^{1} vdu = v\int_{v}^{1} du = v·u\bigg|_{v}^{1} = v(1-v)$
Then, integrate with respect to v: $\int_{0}^{1}\int_{v}^{1} vdudv = \int_{0}^{1} v(1-v) dv = \int_{0}^{1} v-v^2 dv = \frac{1}{2}v^2-\frac{1}{3}v^3\bigg|_{0}^{1} = \frac{1}{2}-\frac{1}{3} = \frac{3}{6}-\frac{2}{6} =\frac{1}{6}.$
Understanding the Region R and the Transformation. Given the transformation: $u = 2x - y$ and $v = 2x + y$, this is a linear transformation, meaning that the region R in the xy-plane will transform into a region R′ in the uv-plane. Since the transformation is linear, lines in the xy-plane are mapped to lines in the uv-plane.
$\star$ $(0,0) \implies u = 2(0)-0 = 0, \quad v = 2(0)+0 = 0 \implies (0,0)$
$\star$ $(1/2, 1) \implies u = 2(1/2)-1 = 0, \quad v = 2(1/2)+1 = 2 \implies (0,2)$
$\star$ $(0,2) \implies u = 2(0)-2 = -2, \quad v = 2(0)+2 = 2 \implies (-2,2)$
$\star$ The new region $R'$ in the $uv$-plane is a triangle with vertices $(0,0)$, $(0,2)$, and $(-2,2)$.
$\star$ The area of the original triangle $R$ (base 2 along the y-axis, height $1/2$) is $\frac{1}{2}(2)(1/2) = 1/2$.
$\star$ The area of the new triangle $R'$ (base 2 along the v-axis, height 2) is $\frac{1}{2}(2)(2) = 2$.
Notice the area scaled by a factor of 4.
Determine the Jacobian.
$\frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} 2 & -1 \\ 2 & 1 \end{vmatrix} = (2)(1) - (-1)(2) = 4$
Using the inverse relationship, $dxdy = \frac{1}{\left\vert{} \frac{\partial(u,v)}{\partial(x,y)} \right\vert{}} dudv = \frac{1}{4} dudv$. This confirms our geometric observation: the area in the $uv$-plane is 4 times larger, so the differential element must be scaled down by $\frac{1}{4}$ to preserve equality.
Set up the integral in the new coordinates.
Notice that $(u)(v) = (2x - y)(2x + y) = 4x^2 - y^2$.
This beautifully simplifies the integrand: $(4x^2 - y^2)^4 = (uv)^4$.
Substituting into the integral: $\iint_{R} (4x^2 - y^2)^4 \,dxdy = \iint_{R'} (uv)^4 \frac{1}{4} \,dudv$
Set Up and Evaluate the Integral in the New Region R′.
Looking at the vertices of $R'$ in the $uv$-plane, $v$ ranges from $0$ to $2$. For a fixed $v$, $u$ ranges from the diagonal line connecting $(0,0)$ and $(-2,2)$ to the vertical line $u=0$. The equation of that diagonal line is $u = -v$. Therefore, the bounds for $u$ are $-v$ to $0$.
$\int_{0}^{2}\int_{-v}^{0} \frac{(uv)^4}{4}dudv = \int_{0}^{2}\int_{-v}^{0} \frac{u^4v^4}{4}dudv = \int_{0}^{2} \frac{v^4}{4} (\int_{-v}^{0} u^4du)dv.$
The inner integral: $\int_{-v}^{0} u^4du = \frac{u^5}{5}\bigg|_{-v}^{0} = -\frac{(-v)^5}{5} = \frac{v^5}{5}$
Substitute this back into the outer integral: $\int_{0}^{2} \frac{v^4}{4} (\int_{-v}^{0} u^4du)dv = \int_{0}^{2} \frac{v^4}{4}\frac{v^5}{5} dv = \int_{0}^{2} \frac{v^9}{20}dv = \frac{1}{20}·\frac{v^{10}}{10}\bigg|_{0}^{2} = \frac{1}{20}\frac{2^{10}}{10} = \frac{1}{20}\frac{1024}{10} = \frac{1024}{200} = 5.12$
If you have ever tried to integrate a function over an awkwardly slanted parallelogram or a skewed region, you know how quickly the algebra can become a nightmare. Evaluating integrals over complex boundaries in the Cartesian $xy$-plane often requires splitting the integral into multiple parts and dealing with messy algebraic bounds.
But what if we could warp space to make the boundaries simple?
By introducing a change of variables, we can transform a difficult region in the $xy$-plane into a simple rectangle or right triangle in a new $uv$-plane. This technique is the multivariable equivalent of $u$-substitution. In this guide, we will walk step-by-step through four complex double integrals, mastering the coordinate transformations and the crucial geometric scaling factor known as the Jacobian.
When we map a region from the $xy$-plane to the $uv$-plane using transformations $x = x(u,v)$ and $y = y(u,v)$, the area of the region stretches or shrinks. The Jacobian Determinant tells us exactly how much a tiny patch of area $du\,dv$ scales when it is mapped to $dx\,dy$.
The Jacobian $J$ is defined as the determinant of the matrix of partial derivatives: $J = \frac{\partial(x, y)}{\partial(u, v)} = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix}$
Because area must always be a positive quantity, we take the absolute value of the Jacobian when rewriting our differential area element: $dx\,dy = \left\vert{} \frac{\partial(x, y)}{\partial(u, v)} \right\vert{} \,du\,dv$
Note: Sometimes it is easier to calculate the inverse transformations $u(x,y)$ and $v(x,y)$. In these cases, we can calculate the inverse Jacobian and use the property: $\frac{\partial(x, y)}{\partial(u, v)} = \left( \frac{\partial(u, v)}{\partial(x, y)} \right)^{-1}$.
Identify the region R over which we are going to integrate.
The side of the parallelogram between $(1, 2)$ and $(3, 4)$ =[The slope m of a line is the ratio of change in y and the change in x, m = $\frac{y_2-y_1}{x_2-x_1}$. A linear equation can be written in the form $y -y_1 = \frac{y_2-y_1}{x_2-x_1} (x -x_1)$]. Therefore, we can determine the boundaries:
$\star$ Between $(1, 2)$ and $(3, 4)$: $y - 2 = 1(x - 1) \implies x - y + 1 = 0$
$\star$ Between $(4, 3)$ and $(6, 5)$: $y - 3 = 1(x - 4) \implies x - y - 1 = 0$
$\star$ Between $(1, 2)$ and $(4, 3)$: $y - 2 = \frac{1}{3}(x - 1) \implies x - 3y + 5 = 0$
$\star$ Between $(3, 4)$ and $(6, 5)$: $y - 4 = \frac{1}{3}(x - 3) \implies x - 3y + 9 = 0$
Define the Change of Variables. The boundary equations naturally suggest our new variables! Notice the recurring expressions $x-y$ and $x-3y$.
Then, we define $u = x - y$ and $v = x - 3y$.
Substituting these into our boundary equations gives us a very simple rectangular region $R'$ in the $uv$-plane:
$\star$ $x - y + 1 = 0 \implies u = -1$
$\star$ $x - y - 1 = 0 \implies u = 1$
$\star$ $x - 3y + 5 = 0 \implies v = -5$
$\star$ $x - 3y + 9 = 0 \implies v = -9$
Our new bounds are simply $-1 \le u \le 1$ and $-9 \le v \le -5$. Thus, the new region $R'$ in the uv-plane is bounded by u ranging from −1 to 1 and v ranging from -9 to -5.
Calculate the Jacobian determinant.
Because we have $u$ and $v$ defined in terms of $x$ and $y$, it is easiest to find the inverse Jacobian first: $\frac{\partial(u, v)}{\partial(x, y)} = \det \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix} = \det \begin{pmatrix} 1 & -1 \\ 1 & -3 \end{pmatrix} = (1)(-3) - (-1)(1) = -2$
Since the actual Jacobian is the reciprocal of this, we have $J = -\frac{1}{2}$. We take the absolute value to find our differential area element: $dx\,dy = \left\vert{} -\frac{1}{2} \right\vert{} \,du\,dv = \frac{1}{2} \,du\,dv$
Set Up the Integral: $\int \int_{R} (x-y)dydx = \int_{-9}^{-5} \int_{-1}^{1} (u) \left(\frac{1}{2}\right) \,du \,dv$
Evaluate the integral
Evaluate the inner integral with respect to u: $\int_{-1}^{1} \frac{u}{2} \,du = \left[ \frac{u^2}{4} \right]_{-1}^{1} = \frac{1}{4} - \frac{1}{4} = 0$
The positive area from $u \gt 0$ cancels the negative area from $u \lt 0$.
Evaluate the outer integral: $\int_{-9}^{-5} 0dv = 0.$ Since the inner integral is 0, the entire integral evaluates to 0.
The Transformation.
To simplify this, we define $u$ and $v$ based on the first two boundaries: $u = y - x$ (i) and $v = y - 2x$ (ii).
Define the region in terms of new variables (Figure 3). The region R is bounded by three lines:
$\star$ Line $y = x \implies u = y - x = x -x = 0$.
$\star$ Line $y = 2x \implies v = y -2x = 2x -2x = 0.$
$\star$ Line $x + y = 2$. First, we realize that $u -v =[\text{i, ii}] (y -x)-(y-2x) = x$, so $x = u-v$ (iii). Substituting $x$ back into our first equation yields $u = y -x \implies y = x + u = (u - v) + u = 2u - v$ (iv).
Now, substitute these into $x + y = 2$: $\text{[iii, iv]} (u-v)+(2u-v) = 2 \implies 3u -2v = 2 \implies v = \frac{3u-2}{2} = \frac{3}{2}u-1.$
Calculate the Jacobian Determinant.
Using our expressions $x = u - v$ and $y = 2u - v$: $J = \frac{\partial(x, y)}{\partial(u, v)} = \det \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} = (1)(-1) - (-1)(2) = 1$
The absolute value is $\vert{}1\vert{} = 1$. This means the area is perfectly preserved on a 1:1 scale during this transformation! So, $dx\,dy = 1 \cdot du\,dv$.
Set Up the Integral.
Notice that $u$ ranges between $0$ and $\frac{2}{3}$. Given a fixed $u$, $v$ ranges between $3u-2v = 2 \iff v = \frac{3u-2}{2}=\frac{3}{2}u-1$ and 0.
$\int \int_{R} (x+y)dxdy = \int_{0}^{\frac{2}{3}}(\int_{\frac{3}{2}u-1}^{0} (3u-2v)dv)du$
Evaluate the Integral.
Evaluating the inner integral with respect to v: $\int_{\frac{3}{2}u-1}^{0} (3u-2v)dv = 3uv-v^2\bigg|_{\frac{3}{2}u-1}^{0} = 0-(3u(\frac{3}{2}u-1)-(\frac{3}{2}u-1)^2) = -3u(\frac{3}{2}u-1)+(\frac{3}{2}u-1)^2 = -\frac{9u^2}{2}+3u+\frac{9u^2}{4}-3u+1 = \frac{-9u^2}{4}+1$
Evaluate the outer integral with respect to u: $\int \int_{R} (x+y)dxdy = \int_{0}^{\frac{2}{3}} (\frac{-9u^2}{4}+1)du = \frac{-9u^3}{4·3}+u = \frac{-3u^3}{4}+u\bigg|_{0}^{\frac{2}{3}} = \frac{-3}{4}·(\frac{2}{3})^3+\frac{2}{3} = \frac{-3}{4}·\frac{8}{27}+\frac{2}{3} = \frac{-2}{9}+ \frac{2}{3} = \frac{-2}{9}+\frac{6}{9} = \frac{4}{9}.$
The Transformation.
This time, we will manually define a transformation that elegantly simplifies the integrand: $x = u - v$ and $y = u + v$.
Identify the region R over which we are going to integrate.
$\star$ Boundary $y = x \implies u + v = u -v \implies 2v = 0 \implies v = 0$.
$\star$ Boundary $y = 3x \implies u + v = 3(u -v) \implies u + v = 3u -3v \implies 4v = 2u \implies v = \frac{u}{2}$.
$\star$ Boundary $x + y = 4 \implies u + v + u - v = 4 \implies 2u = 4 \implies u = 2.$
Refer to Figure i for a visual representation and aid in understanding it. Therefore, S is the region bounded by $v = 0, v = \frac{u}{2}$, and $u = 2$.
Calculate the Jacobian Determinant. Using our transformation $x = u - v$ and $y = u + v$: $J = \frac{\partial(x, y)}{\partial(u, v)} = \det \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = (1)(1) - (-1)(1) = 2$
Thus, $dx\,dy = \vert{}2\vert{} \,du\,dv = 2 \,du\,dv$.
Set Up the Integral. Notice that $u$ ranges between $0$ and $2$. Given a fixed $u$, $v$ ranges between $0$ and $\frac{u}{2}$.
Our original integrand $x+y$ simplifies beautifully to $(u-v) + (u+v) = 2u$.
$\int \int_{R} (x+y)dxdy = \int_{0}^{2}(\int_{0}^{\frac{u}{2}} (4u)dv)du = 4\int_{0}^{2}(\int_{0}^{\frac{u}{2}} u dv)du$
Evaluate the Integral
Evaluating the inner integral with respect to v: $\int_{0}^{\frac{u}{2}} udv = uv\bigg|_{0}^{\frac{u}{2}} = \frac{u^2}{2}$
Evaluating the outer integral with respect to u: $\int \int_{R} (x+y)dxdy = \frac{4}{2}\int_{0}^{2}u^2du = 2\frac{u^3}{3}\bigg|_{0}^{2} = 2\frac{2^3}{3} = \frac{16}{3}$.
Identify the region R over which we are going to integrate. R is the region bounded by $x = 1; x = 2; y = x + 2$ and $y = x + 3$.
$\star$ $y = x + 2$ (lower boundary) $\implies u + v = u + 2 \implies v = 2$.
$\star$ $y = x + 3$ (upper boundary) $\implies u + v = u + 3 \implies v = 3$.
$\star$ $x = 1$ (left boundary) $\implies u = 1$.
$\star$ $x = 2$ (right boundary) $\implies u = 2$.
Thus, the transformed region R is a perfect rectangle bounded by $1 \le u \le 2$ and $2 \le v \le 3$.
Expressing $\frac{1}{\sqrt{xy-x^2}}$ in Terms of u and v .
Substitute $x=u$ and $y=u+v$ into the denominator: $xy - x^2 = x(y-x) = u(v) = u \cdot v$
Our integrand becomes exactly $\frac{1}{\sqrt{uv}}$.
Calculate the Jacobian Determinant. $J = \frac{\partial(x, y)}{\partial(u, v)} = \det \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = (1)(1) - (0)(1) = 1$
So, $dx\,dy = 1 \cdot du\,dv$.
Set Up the Integral. Notice that u ranges between 1 and 2 and v ranges between 2 and 3. $\int_{1}^{2}(\int_{2}^{3} (\frac{1}{\sqrt{uv}})dv)du = \int_{1}^{2}(\int_{2}^{3} (u^{\frac{-1}{2}}v^{\frac{-1}{2}})dv)du$
Evaluate the Integral.
Evaluating the inner integral with respect to v: $\int_{2}^{3} (u^{\frac{-1}{2}}v^{\frac{-1}{2}})dv = u^{\frac{-1}{2}}\int_{2}^{3} v^{\frac{-1}{2}}dv = u^{\frac{-1}{2}}2v^{\frac{1}{2}}\bigg|_{2}^{3}= 2(\sqrt{3}-\sqrt{2})u^{\frac{-1}{2}}$
Evaluating the outer integral with respect to u: $\int_{1}^{2}2(\sqrt{3}-\sqrt{2})u^{\frac{-1}{2}}du = 4(\sqrt{3}-\sqrt{2})u^{\frac{1}{2}}\bigg|_{1}^{2}$
$ = 4(\sqrt{3}-\sqrt{2})u^{\frac{1}{2}}\bigg|_{1}^{2} = 4(\sqrt{3}-\sqrt{2})(\sqrt{2}-1) = 4\sqrt{2}\sqrt{3}-8-4\sqrt{3}+4\sqrt{2}$.
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