The things you own end up owning you, Fight Club
A function of two variables f: ℝ x ℝ → ℝ assigns to each ordered pair in its domain a unique real number, e.g., Area = $\frac{1}{2}b·h$, z = f(x, y) = 2x + 3y, f(x, y) = x^{2} + y^{2}, e^{x+y}, etc.
Partial derivatives are derivatives of a function of multiple variables, say f(x_{1}, x_{2}, ···, x_{n}) with respect to one of those variables, with the other variables held constant. They measure how the function changes as one variable changes, while keeping the others constant. $\frac{\partial f}{\partial x}(x_0, y_0) = \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x, y_0)-f(x_0, y_0)}{\Delta x}$
Geometrically, the partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ at a point (x_{0}, y_{0}) can be interpreted as the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane y = y_{0}. Similarly, $\frac{\partial f}{\partial y}(x_0, y_0)$ represents the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane x = x_{0}
$\frac{dw}{ds}\bigg|_{\vec{u}} = ∇w·\vec{u} = |∇w|·|\vec{u}|cos(θ) = |∇w|·cos(θ)$ where θ is the angle between the gradient and the given unit vector.
The gradient is the direction in which the function increases fastest (the direction of the steepest ascent) at a given point, and |∇w| = $\frac{dw}{ds}\bigg|_{\vec{u}=dir(∇w)}$.
The directional derivative is minimized when cos(θ) = -1 ↭ θ = 180 ↭ $\vec{u}$ is in the opposite direction of the gradient ∇w. Futhermore, $\frac{dw}{ds}\bigg|_{\vec{u}} = 0$ ↭ cos(θ) = 0 ↭ θ = 90° ↭ $\vec{u}$ ⊥ ∇w.
The Chain Rule. Let w = f(x, y, z) where x = x(t), y = y(t), and z = z(t) are functions of another variable t, then the derivate of f with respect to t is given by $\frac{dw}{dt} = f_x\frac{dx}{dt} +f_y\frac{dy}{dt}+f_z\frac{dz}{dt}, \frac{dw}{dt} = w_x\frac{dx}{dt} +w_y\frac{dy}{dt}+w_z\frac{dz}{dt} = ∇w·\frac{d\vec{r}}{dt}~$ where ∇w = ⟨w_{x}, w_{y}, w_{z}⟩ and $\frac{d\vec{r}}{dt}=⟨\frac{d\vec{x}}{dt}, \frac{d\vec{y}}{dt}, \frac{d\vec{z}}{dt}⟩$
The gradient vector is defined as follows: ∇w = ⟨w_{x}, w_{y}, w_{z}⟩ = $⟨\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z}⟩$. And $\frac{d\vec{r}}{dt}$ is the vector of derivatives of x, y, and z with respect to t: $\frac{d\vec{r}}{dt} = ⟨\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}⟩$.
The gradient vector is a vector that points in the direction of the steepest increase of the function at a given point. It is perpendicular (orthogonal) to the level surfaces of the function. This means that the gradient points in the direction of the steepest ascent or increase, and there is no change in the function’s value along the level surface, making the gradient perpendicular to it.
Lagrange multipliers are a strategy for finding the local maxima and minima of a function, say f(x, y, z), where the variables x, y, and z are not independent, but subject to an equality constraint, g(x, y, z) = c. This method uses the fact that any point where f(x, y, z) reaches a constraint extremum (maximum or minimum), the gradient of f is parallel to the gradient of g. Formally, this relationship is expressed as ∇f = λ∇g where λ is the Lagrange multiplier.
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] x [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: $\int \int_R f(x, y)dA = \int_{a}^{b}\int_{c}^{d} f(x, y)dydx = \int_{c}^{d}\int_{b}^{a} f(x, y)dxdy$
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 more general transformation, the integral can often be simplified, making the calculation more manageable.
Let’s explore how this technique works by calculating the area of an ellipse. An ellipse with with semi-major axis a and b is described by the equation $(\frac{x}{a})^2 + (\frac{y}{b})^2 = 1$. If r = a = b, this equation describes a circle.
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 ≤ 1} dxdy$
In this particular case, the change of variables was pretty easy and transformed the problem from integrating over an ellipse to integrating over a circle—a much simpler shape. This approach is particularly useful when dealing with complex regions in the xy-plane, where a direct approach might be cumbersome or impossible.
The previous change of variables was pretty easy, but this is not always the case. When we change variables in a double integral, we need to account for how the area element dA = dxdy in the original coordinates transforms into a new area element dA’ in the new coordinates. This transformation involves the Jacobian determinant, which provides a scaling factor that adjusts the area element for the transformation.
Consider a linear transformation defined by the equations: 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 ΔA rectangle in the xy-plane. Since the transformations u = 3x −2y and v = x + y are linear, this rectangle is mapped to a parallelogram ΔA′ in the uv-plane. The important property of linear transformations is that they scale areas by a constant factor that is independent on the choice (specific location) of the rectangle (Figure 1 and 2).
In geometric terms, if we consider two vectors representing the sides of the rectangle ΔA in the xy-plane, their images under the transformation will also be vectors that define the parallelogram ΔA′ in the uv-plane.
The area of the parallelogram ΔA′ can be calculated using the determinant of the matrix formed by the transformation. We will try to understand this with the unit rectangle (Figure 3).
Recall that for any two vectors $\vec{u} = ⟨u_1, u_2⟩$ and $\vec{v} = ⟨v_1, v_2⟩$ in a 2-dimensional space, the area of the parallelogram spanned by those vectors is given by the absolute value of the determinant of the matrix: Area = $|\begin{smallmatrix}u_1 & v_1\\\ u_2 & v_2\end{smallmatrix}|$.
For the transformation u = 3x −2y and v = x + y, the Jacobian matrix J is: A’ = $|\begin{smallmatrix}\frac{∂u}{∂x} & \frac{∂u}{∂y}\\ \frac{∂v}{∂x} & \frac{∂v}{∂y}\end{smallmatrix}| = |\begin{smallmatrix}3 & 1\\ -2 & 1\end{smallmatrix}|$.
The determinant of this matrix is: det(J) = 3⋅1 −(−2)⋅1 = 3 + 2 = 5. More importantly, for any arbitrary rectangle, the area scaling factor is always 5, so dA' = 5dA or dudv = 5dxdy.
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 R, 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)}|·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 = (x, y), which define a transformation from the xy-plane to the uv-plane. Small changes Δx and Δy in the xy-plane lead to corresponding changes Δu and Δv in the uv-plane, which can be approximated as: $Δu ≈ \frac{∂u}{∂x}Δx + \frac{∂u}{∂y}Δy$ = u_{x}Δx + u_{y}Δy where $\frac{∂u}{∂x}, \frac{∂u}{∂y}$ 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 Δv ≈ v_{x}Δx + v_{y}Δy.
This relationships can be written or expressed in matrix form: $|\begin{smallmatrix}Δu\\ Δv\end{smallmatrix}| = |\begin{smallmatrix}u_x & u_y\\ v_x & v_y\end{smallmatrix}||\begin{smallmatrix}Δx\\ Δy\end{smallmatrix}|$ (Figure 4).
The determinant of this Jacobian matrix J gives the scaling factor for the area element: det(J) = $det(\frac{∂(u,v)}{∂(x,y)}) = |\begin{smallmatrix}u_x & u_y\\ v_x & v_y\end{smallmatrix}|⇒dA^’=∣J∣dA ⇒ dudv = |J|dxdy= |\begin{smallmatrix}u_x & u_y\\ v_x & v_y\end{smallmatrix}|dxdy$.
Thus, the differential area elements transform as: dA′ =∣det(J)∣dA or dudv = ∣det(J)∣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: $\iint_R f(x, y)dA = \iint_S f(x(u,v), y(u, v))·|\frac{1}{det(J)}|dudv = \iint_S f(x(u,v), y(u, v))·|\frac{1}{\begin{smallmatrix}u_x & u_y\\\ v_x & v_y\end{smallmatrix}}|dudv$.
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 $|\frac{1}{det(J)}| = \frac{1}{|\begin{smallmatrix}u_x & u_y\\ v_x & v_y\end{smallmatrix}|}$ is the reciprocal of the determinant of the Jacobian matrix of the transformation. It is a factor that adjust the area element for the change in the scale cause 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: $\frac{∂(u,v)}{∂(x,y)}·\frac{∂(x,y)}{∂(u,v)} = 1$
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, θ) 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, θ) is given by the following equations: x = rcos(θ), y = rsin(θ). Here, r represents the radial distance from the origin, and θ 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 (x, y) → (r, θ) is constructed by taking the partial derivatives of x and y with respect to r and θ: $J = \frac{∂(x, y)}{∂(r, θ)} = (\begin{smallmatrix}\frac{∂x}{∂r} & \frac{∂x}{∂θ}\\ \frac{∂y}{∂r} & \frac{∂y}{∂θ}\end{smallmatrix}) = (\begin{smallmatrix}x_r & x_θ\\ y_r & y_θ\end{smallmatrix}) = (\begin{smallmatrix}cos(θ) & -rsin(θ)\\ sin(θ) & rcos(θ)\end{smallmatrix}).$
The Jacobian determinant det(J) is: det(J) = $rcos^2(θ)+rsin^2(θ) = r(cos^2(θ)+sin^2(θ)) = r·1 = r$ ⇒ dxdy = |r|drdθ =[r is always positive] rdrdθ. The Jacobian determinant for this transformation is r, so the area element dA in polar coordinates becomes rdrdθ (dA = rdrdθ). 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 1/r.
Steps for Changing variables.
Identify the new coordinates: u = x, v = xy
Determine the Jacobian determinant, $det(J) = \frac{∂(u, v)}{∂(x, y)} = |\begin{smallmatrix}\frac{∂u}{∂x} & \frac{∂u}{∂y}\\ \frac{∂v}{∂x} & \frac{∂v}{∂y}\end{smallmatrix}| = |\begin{smallmatrix}1 & 0\\ y & x\end{smallmatrix}| = x$. This determinant is crucial as it gives us the scaling factor for the area element in the new coordinates.
Setup 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). Here, the limits of integration for u and v are determined based on the original region of integration in the xy-plane.
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.
The area of region R in the xy-plane is given as: Area(R) = $\frac{1}{2}·2·\frac{1}{2}=\frac{1}{2}$. The area of the transformed region R′ in the uv-plane is: Area(R’) = $\frac{1}{2}·2·2 = 2⇒ \frac{1}{2}·4 = 2$.
Since the transformation changes the area by a factor of 4, the differential area elements are related by: dA′ = dudv = 4dxdy.
Determine the Jacobian determinant.
The Jacobian matrix J for the transformation is computed by taking the partial derivatives of u and v with respect to x and y:
$det(J) = \frac{∂(u, v)}{∂(x, y)} = |\begin{smallmatrix}\frac{∂u}{∂x} & \frac{∂u}{∂y}\\ \frac{∂v}{∂x} & \frac{∂v}{∂y}\end{smallmatrix}| = |\begin{smallmatrix}2 & -1\\ 2 & 1\end{smallmatrix}| = (2)(1)−(−1)(2) = 2-(-2)=2+2=4.$
Thus, the scaling factor between the areas in the xy-plane and the uv-plane is 4, confirming that dA′ = 4dxdy.
Setup the integral in the new coordinates.
The differential area dA’ = dudv = 4·dxdy. Given the relationship between u, v and the original integrand, observe that: u·v = (2x -y)(2x + y) = 4x^2-y^2. This simplifies the original integrand: $(4x^2-y^2)^4 = (uv)^4.$
Substituting into the integral, we obtain:
$\int \int_{R} (4x^2-y^2)^4dxdy = \int \int_{R} \frac{(uv)^4}{4}dudv$
Set Up and Evaluate the Integral in the New Region R′.
v ranges from 0 to 2. This range is determined by the transformation and the limits of the original region R in the xy-plane. Fixing v, the lower boundary for u is u =−v (when y=1). The upper boundary for u is u = 0 (when y=0). In other words, for a fixed v v, u ranges from -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: $\int_{0}^{2} \frac{v^4}{4} (\int_{-v}^{0} u^4du)dv = \int_{0}^{2} \frac{v^4}{4}\frac{v^5}{5} dv = \int_{0}^{2} \frac{v^9}{20}dv = \frac{1}{20}·\frac{v^{10}}{10}\bigg|_{0}^{2} = \frac{1}{20}\frac{2^{10}}{10} = \frac{1}{20}\frac{1024}{10} = \frac{1024}{200} = 5.12$
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 $y -y_1 = \frac{y_2-y_1}{x_2-x_1} (x -x_1)$] $y -2 = \frac{4-2}{3-1}(x-1)↭ y -2 = 1·(x-1)↭ x-y+1=0$.
Similarly, we can compute the four sides of the parallelogram, namely x −y +1 = 0, x −y −1=0, x −3y +5 = 0 and x −3y +9 = 0.
Change of variables u = x−y and v = x−3y. Calculate the Jacobian determinant. The Jacobian determinant is crucial for transforming the differential area element dxdy into dudv.
$det(J) = \frac{∂(u, v)}{∂(x, y)} = |\begin{smallmatrix}\frac{∂u}{∂x} & \frac{∂u}{∂y}\\ \frac{∂v}{∂x} & \frac{∂v}{∂y}\end{smallmatrix}| = |\begin{smallmatrix}1 & -1\\ 1 & -3\end{smallmatrix}| = 1·(-3)-1(-1)=-3+1 = -2$. So, the Jacobian determinant is det(J) = −2.
Transform the Boundaries. We now transform the boundary equations of the parallelogram into the uv-plane: (x -y = u, x -3y = v)
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. By symmetry, we could also express the region with we get the same results by u ranging from −1 to 1 and v ranging from 5 to 9. This is because the transformation effectively mirrors the region about the origin.
Set Up the Integral: $\int \int_{R} (x-y)dydx = \int_{-9}^{-5}(\int_{-1}^{1} \frac{u}{det(J)}du) dv = \int_{-9}^{-5}(\int_{-1}^{1} \frac{u}{-2}du) dv$
Evaluate the integral
Evaluate the inner integral with respect to u: $\int_{-1}^{1} \frac{u}{-2}du = \frac{-1}{2}\frac{u^2}{2}\bigg|_{-1}^{1} = \frac{-1}{2}(\frac{1}{2}-\frac{1}{2})= 0$
Evaluate the outer integral: $\int_{-9}^{-5}(\int_{-1}^{1} \frac{u}{2}du) dv = \int_{-9}^{5} 0dv = 0.$ Since the inner integral is 0, the entire integral evaluates to 0.
Define the region in terms of new variables (Figure 3). The region R is bounded by three lines:
x + y = 2 ↭ [iii, iv] (u-v)+(2u-v) = 2 ↭ 3u -2v = 2 ⇒v = $\frac{3u-2}{2} = \frac{3}{2}u-1.$
Calculate the Jacobian Determinant. The Jacobian determinant det(J) for the transformation y -x = u (i), y -2x = v(ii) is calculated as: $det(J) = \frac{∂(u, v)}{∂(x, y)} = |\begin{smallmatrix}\frac{∂u}{∂x} & \frac{∂u}{∂y}\\ \frac{∂v}{∂x} & \frac{∂v}{∂y}\end{smallmatrix}| = |\begin{smallmatrix}-1 & 1\\ -2 & 1\end{smallmatrix}| = (-1)·1-1(-2)=-1+2 = 1$
Set Up the Integral.
Notice that u ranges between 0 and $\frac{2}{3}$. Given a fixed u, v ranges between 3u-2v = 2(↭v = $\frac{3u-2}{2}=\frac{3}{2}u-1$) and 0. x+y =[(iii), (iv)] u -v + (2u -v) = 3u -2v.
$\int \int_{R} (x+y)dxdy =[iii, iv] \int_{0}^{\frac{2}{3}}(\int_{\frac{3}{2}u-1}^{0} (3u-2v)dv)du$
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 integrate with respect to u:
$\int \int_{R} (x+y)dxdy = \int_{0}^{\frac{2}{3}} (\frac{-9u^2}{4}+1)du = \frac{-9u^3}{4·3}+u = \frac{-3u^3}{4}+u\bigg|_{0}^{\frac{2}{3}} = \frac{-3}{4}·(\frac{2}{3})^3+\frac{2}{3} = \frac{-3}{4}·\frac{8}{27}+\frac{2}{3} = \frac{-2}{9}+ \frac{2}{3} = \frac{-2}{9}+\frac{6}{9} = \frac{4}{9}.$
Identify the region R over which we are going to integrate.
Expressing x + y in Terms of u and v:
x = u -v (i), y = u + v (ii), (i+ii) x + y = 2u ⇒ $u = \frac{1}{2}(x+y)$
x = u -v (i), y = u + v (ii), (i-ii) x - y = -2v ⇒ $v = \frac{-1}{2}(x-y)$
x + y = (u - v) + (u + v) = 2u.
Calculate the Jacobian Determinant. The Jacobian determinant det(J) for the transformation $u = \frac{1}{2}(x+y)$ (i), $v = \frac{-1}{2}(x-y)$ (ii) is calculated as: $det(J) = \frac{∂(u, v)}{∂(x, y)} = |\begin{smallmatrix}\frac{∂u}{∂x} & \frac{∂u}{∂y}\\ \frac{∂v}{∂x} & \frac{∂v}{∂y}\end{smallmatrix}| = |\begin{smallmatrix}\frac{1}{2} & \frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2}\end{smallmatrix}| = \frac{1}{4}+\frac{1}{4} = \frac{1}{2}$
Set Up the Integral.
Notice that u ranges between 0 and 2. Given a fixed u, v ranges between 0 amd $\frac{u}{2}$.
$\int \int_{R} (x+y)dxdy =\int_{0}^{2}(\int_{0}^{\frac{u}{2}} \frac{1}{|det(J)|}(2u)dv)du = \int_{0}^{2}(\int_{0}^{\frac{u}{2}} (4u)dv)du = 4\int_{0}^{2}(\int_{0}^{\frac{u}{2}} (u)dv)du$
Evaluating the inner integral with respect to v: $\int_{0}^{\frac{u}{2}} (u)dv = uv\bigg|_{0}^{\frac{u}{2}} = \frac{u^2}{2}$
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.
Thus, the transformed region R is bounded by 1 ≤ u ≤ 2 and 2 ≤ v ≤ 3.
Expressing $\frac{1}{\sqrt{xy-x*2}}$ in Terms of u and v: x = u (i) ⇒ u = x. y = u + v (ii) ⇒ v = y -u =(i) y -x. $\frac{1}{\sqrt{xy-x*2}} = \frac{1}{\sqrt{u(u+v)-u^2}} = \frac{1}{\sqrt{u^2+uv-u^2}} = \frac{1}{\sqrt{uv}}$
Calculate the Jacobian Determinant. The Jacobian determinant det(J) for the transformation $u = \frac{1}{2}(x+y)$ (i), $v = \frac{-1}{2}(x-y)$ (ii) is calculated as: $det(J) = \frac{∂(u, v)}{∂(x, y)} = |\begin{smallmatrix}\frac{∂u}{∂x} & \frac{∂u}{∂y}\\ \frac{∂v}{∂x} & \frac{∂v}{∂y}\end{smallmatrix}| = |\begin{smallmatrix}1 & 0\\ -1 & 1\end{smallmatrix}| = 1·1-0·(-1) = 1$
Set Up the Integral.
Notice that u ranges between 1 and 2 and v ranges between 2 and 3.
$\int \int_{R} (x+y)dxdy = \int_{1}^{2}(\int_{2}^{3} \frac{1}{|det(J)|}(\frac{1}{\sqrt{uv}})dv)du = \int_{1}^{2}(\int_{2}^{3} (\frac{1}{\sqrt{uv}})dv)du = \int_{1}^{2}(\int_{2}^{3} (u^{\frac{-1}{2}}v^{\frac{-1}{2}})dv)du$
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: $\int \int_{R} (x+y)dxdy = \int_{1}^{2}(\int_{2}^{3} \frac{1}{|det(J)|}(\frac{1}{\sqrt{uv}})dv)du = \int_{1}^{2}2(\sqrt{3}-\sqrt{2})u^{\frac{-1}{2}}du = 4(\sqrt{3}-\sqrt{2})u^{\frac{1}{2}}\bigg|_{1}^{2}$
$ = 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}$.