Change of Variables

Sometimes, it is hard to solve an integral using the given variables like x and y. So, we change the variables to new ones, like u and v, which makes the integral easier. We write x and y in terms of u and v. Then, we also change the limits of integration. This helps us solve the integral more easily.

Change of variable is a method used to simplify an integral by replacing the original variable with new variable.

The Jacobian method is used to perform a change of variable in a double integral. It often changes the variable from (x, y) to a new pair of variables (u, v) to simplify the limits.

The jacobian of the transformation x = g(u, v) , y = h(u,v)

\frac{\partial(x,y)}{\partial(u,v)} =\begin{vmatrix}\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\\dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}\end{vmatrix}

Change of Variable for a Double Integral

Suppose that we want to integrate f(x, y) over the region R under the transformation x = g(u, v), y = h(u, v), the region becomes S, and the integral becomes:

\iint_{R} f(x,y)\, dA = \iint_{S} f\big(g(u,v),\, h(u,v)\big) \left| \frac{\partial(x,y)}{\partial(u,v)} \right| \, d\bar{A}

Solved Question on Change of Variables

Question 1: Evaluate I = \int_0^1 \int_0^{\sqrt{1-x^2}} f(x^2+y^2)\, dy\, dx using the change of variables.

Solution:

Identify the region:
The limits are:

• x: 0 → 1
• y: 0 → 1​

This describes the quarter circle of radius 1 in the first quadrant.

R = {(x,y):x² + y² ≤ 1, x ≥ 0, y ≥ 0}

Since f(x² + y²) depends on x²+y², polar coordinates are natural.

Let x = rcos⁡θ, y = rsin⁡θ

Jacobian:

∂(x,y)/∂(r, θ) = r

• x ≥ 0, y ≥ 0  ⟹  θ ∈ [0, π/2]
• x² + y² ≤ 1  ⟹  r ∈ [0, 1]

I = \int_{0}^{\pi/2} \int_{0}^{1} f(r^2)\, r \, dr \, d\theta

I = \left( \int_{0}^{\pi/2} d\theta \right)\left( \int_{0}^{1} r f(r^2)\, dr \right) = \frac{\pi}{2} \int_{0}^{1} r f(r^2)\, dr

Substitue u = r², du = 2r dr

r dr = 1/2 du

I = \frac{\pi}{2} \int_{0}^{1} r f(r^2)\, dr

I = \frac{\pi}{4} \int_{0}^{1} f(u)\, du

\int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} f(x^2+y^2)\, dy\, dx = \frac{\pi}{4} \int_{0}^{1} f(u)\, du

Question 2: Evaluate I = \iint_R (x+y)\, dA where (R) is the parallelogram with vertices (0, 0), (2, 1), (3, 3), (1, 2), using the transformation x = u + v, y = u - v.

Solution:

Transformation and Jacobian:

x = u + v, y = u - v implies that

u = \frac{x+y}{2}, \quad v = \frac{x-y}{2}

\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -2\quad \Rightarrow \quad \left|\frac{\partial(x,y)}{\partial(u,v)}\right| = 2

Transform the integrand: x + y = (u + v) + (u - v) = 2u

Transform the region (R): Vertices in (uv)-plane:

(0, 0) → (0, 0), (2, 1) →\frac{3}{2}, \frac{1}{2}

So, the rectangle in uv-plane:

0 ≤ u ≤ 3, −12 ≤ v ≤ 120

I = ∬_S​(2u)⋅2dudv = ∬_S​4u du dv​

I = \int_0^3 \int_{-1/2}^{1/2} 4u \, dv \, du= \int_0^3 4u \cdot 1 \, du= 4 \int_0^3 u \, du= 4 \cdot \frac{9}{2} = 18

I = 18

Unsolved Questions on Change of Variables

Question 1: Let R be the parallelogram with vertices (0, 0), (2, 0), (3, 1), (1, 1). Evaluate I=\iint_{R} (3x+y)\,dA using the linear change of variables u = x - y, v = x + 2y .

Question 2: Let the transformation S = { (u,v): 1 ≤ u ≤ 2,  0 ≤ v ≤ 1 }. Evaluate I=\iint_{R} y\,e^{x/y}\,dA where R = T(S) by changing variables to u, v.

Question 3: Evaluate the double integral of 1/(1 + x² + y²) over the annular sector with radius from 1 to 2 and angle from π/4 to π/2 using polar coordinates.

Question 4: Map the rectangle 0 ≤ u ≤ 1,  0 ≤ v ≤ 20 by x = 2u + v, y = u + 3vx = 2u + v, and evaluate the integral of 3x − 2y over the resulting region (use the Jacobian).