What is a double integral?

A double integral is a way to integrate a function of two variables over a two-dimensional region. It is used to calculate areas, volumes, and other quantities that depend on two variables.

How do you set up a double integral?

To set up a double integral, you need to determine the limits of integration for both variables and the integrand (the function to be integrated). The order of integration (whether you integrate with respect to x first or y first) depends on the region of integration and can sometimes simplify the calculation.

What is the geometric interpretation of a double integral?

The geometric interpretation of a double integral is often the volume under a surface in three-dimensional space. If the function represents a height above the xy-plane, the double integral calculates the volume between the surface and the plane over the specified region.

How do you change the order of integration in a double integral?

To change the order of integration, you need to redraw the region of integration and re-evaluate the limits for each variable. This can often simplify the integral, especially if the original order results in complicated limits.

What is the difference between iterated integrals and double integrals?

An iterated integral is an integral performed in a sequence, one variable at a time. A double integral is the combined process of iterating over both variables to integrate over a two-dimensional region. Essentially, a double integral is computed as an iterated integral.

How do you evaluate a double integral over a non-rectangular region?

To evaluate a double integral over a non-rectangular region, you need to carefully set the limits of integration to match the boundaries of the region. This often involves using piecewise functions or splitting the region into simpler subregions.

How do you use polar coordinates to evaluate a double integral?

To use polar coordinates, you transform the x and y coordinates into r (radius) and θ (angle). The double integral is then expressed in terms of r and θ, and the integrand is multiplied by the Jacobian determinant r. The limits of integration are also adjusted to reflect the new coordinates.

Can double integrals be used to find areas?

Yes, double integrals can be used to find areas by integrating a constant function (usually 1) over the region.

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JEE Maths

Double Integral

Double integration is a mathematical process used to compute the integral of a function over a two-dimensional region. It involves two successive integrations and is often used to find areas, volumes, and other quantities that extend over a plane. Formally, if f(x, y) is a function defined over a region R in the xy -plane, the double integral of f over R is denoted as:

$\iint_{R} f (x, y) dA$

where dA represents the differential area element. The double integral can be evaluated iteratively by integrating f with respect to one variable while treating the other variable as a constant, and then integrating the resulting expression with respect to the second variable.

1.0Double Integral

An integral in which the integrand is integrated twice is a double integral,

$\iint f (x) (dx)^{2}$

The integration is performed from the inside out.

Example 1: Find $\int_{- 1}^{3} \int_{0}^{2} 2 x (d x)^{2} .$

Solution: $\int 2 x d x = x^{2} + C_{1}; ∣_{0}^{2} x^{2} = 4$

$\int 4 dx = 4 x + C_{2}; ∣_{- 1}^{3} 4 x = 16$

Hence, $\int_{- 1}^{3} \int_{0}^{2} 2 x d x^{2} = 16$

Functions of more than one variable may be integrated with respect to one variable at a time while the other variables are held constant, reversing the process of partial differentiation.

Example 2: Find $\int_{0}^{3} \int_{1}^{2} (4 y^{3} + 2 x) d x d y$

Solution: Hold y constant and integrate with respect to x:

∫ (4y³+ 2x) dx = 4y³ x + x² + C;

$[4 y^{3} x + x^{2}]_{x = 1}^{x = 2}$ = (4y³. 2 + 2²) – (4y³+1²) = 4y³+ 3

Integrate with respect to y:

∫(4y³+3) dy = y⁴ + 3y + C;

$[y^{4} + 3 y]_{0}^{3} = 81 + 9 = 90.$

Hence,

$\int_{0}^{3} \int_{1}^{2} (4 y^{3} + 2 x) d x d y = 90$

Example 3: Find $\int_{1}^{3} \int_{0}^{2 y} 3 x^{2} d x d y$ .

Solution: $\int_{1}^{3} \int_{0}^{2 y} 3 x^{2} d x d y = \int_{1}^{3} [x^{3}]_{x = 0}^{x = 2 y} d y = \int_{1}^{3} 8 y^{3} d y = ∣_{1}^{3} 2 y^{4} = 160$

2.0Geometrical Interpretation of the Double Integral

We are familiar with the geometrical interpretation of the equation y = f(x) as a curve in the two-dimensional x, y- plane, and of the integral $\int_{a}^{b} f (x) d x$ as an area between the curve and the x- axis.

Similarly, while the equation z = f(x, y) defines a surface in the three-dimensional x, y, z - space , the double integral of a continuous function of two variables $\int_{a}^{b} \int_{c}^{d} f (x, y) d x d y$ , may be interpreted as a volume between the surface z= f(x, y) and the x, y-plane.

Graph showing geometrical interpretation of the double integral

In the graph, the rectangular area ΔA_i in the x, y-plane is projected on the surface z = f (x, y).

The quantity ΔA_i = Δx_i Δy_iis the area of the bottom surface of a column whose top surface is part of the surface f (x_i, yi).

The smaller the area ΔA_i, the closer the volume of the column is to that of a parallelepiped measuring f(x_i, y_i) ΔA_i.

If the domain D, consisting of (x, y) with c ≤ x ≤ d, a ≤ y ≤ b, is divided into an increasingly greater number of rectangles so that ΔA_i tends to 0, then the volume between the surface z = f(x, y) and D equals then the sum of all parallelepipeds measuring f(x_i, y_i) Δx_i Δy_i ; thus,

$V = lim_{n \to \infty} \sum_{i = 1}^{n} f (x_{i}, y_{i}) Δ x_{i} Δ y_{i} = \int_{a}^{b} \int_{c}^{d} f (x, y) d x d y$ .

3.0Solved Examples on Double Integral

Example 1: Evaluate the double integral $\iint_{R} (x^{2} + y^{2}) d A$ over the region R where $0 \leq x \leq 1$ and $0 \leq y \leq 1$ :

Step 1: Set Up the Double Integral

$\iint_{R} (x^{2} + y^{2}) d A = \int_{0}^{1} (\int_{0}^{1} (x^{2} + y^{2}) d y) d x$

Step 2: Integrate with Respect to y

$\int_{0}^{1} (x^{2} + y^{2}) d y = [x^{2} y + \frac{y ^{3}}{3}]_{0}^{1} = x^{2} (1) + \frac{1 ^{3}}{3} - 0 = x^{2} + \frac{1}{3}$

Step 3: Integrate with Respect to x

$\int_{0}^{1} (x^{2} + \frac{1}{3}) d x = [\frac{x ^{3}}{3} + \frac{x}{3}]_{0}^{1} = (\frac{1}{3} + \frac{1}{3}) - 0 = \frac{2}{3}$

Thus, the value of the double integral is $\frac{2}{3}$ .

Example 2: Evaluate the double integral $\iint_{R} x e^{x y} d A$ over the region R where $0 \leq x \leq 1$ and $0 \leq y \leq 1$ :

Step 1: Set Up the Double Integral

$\iint_{R} x e^{x y} d A = \int_{0}^{1} (\int_{0}^{1} x e^{x y} d y) d x$

Step 2: Integrate with Respect to y

$\int_{0}^{1} x e^{x y} d y = x \int_{0}^{1} e^{x y} d y = x [\frac{e ^{x y}}{x}]_{0}^{1} = x (\frac{e ^{x} - 1}{x}) = e^{x} - 1$

Step 3: Integrate with Respect to x

$\int_{0}^{1} (e^{x} - 1) d x = [e^{x} - x]_{0}^{1} = (e^{1} - 1) - (1 - 0) = e - 2$

Thus, the value of the double integral is e – 2.

Example 3: Evaluate the double integral $\iint_{R} (3 x + 4 y) d A$ over the triangular region R with vertices at (0,0), (1,0), and (0,1):

Step 1: Set Up the Double Integral

The region R is bounded by y = 0, x = 0, and y = 1 – x . Therefore, we integrate with respect to y first:

$\iint_{R} (3 x + 4 y) d A = \int_{0}^{1} (\int_{0}^{1 - x} (3 x + 4 y) d y) d x$

Step 2: Integrate with Respect to y

$\int_{0}^{1 - x} (3 x + 4 y) d y = [3 x y + 2 y^{2}]_{0}^{1 - x}$

=3 x(1-x)+2(1-x)^2-0

=3 x-3 x^2+2-4 x+2 x^2

=-x^2-x+2

Step 3: Integrate with Respect to x

$\int_{0}^{1} (- x^{2} - x + 2) d x = [\frac{- x ^{3}}{3} - \frac{x ^{2}}{2} + 2 x]_{0}^{1}$

$= (\frac{- 1}{3} - \frac{1}{2} + 2) - 0$

$= \frac{7}{6}$

Thus, the value of the double integral is $\frac{7}{6}$ .

Example 4:

Evaluate the double integral $\iint_{R} x y d A$ over the region R where $0 \leq x \leq 2$ and $0 \leq y \leq x$ :

Step 1: Set Up the Double Integral

$\iint_{R} x y d A = \int_{0}^{2} (\int_{0}^{x} x y d y) d x$

Step 2: Integrate with Respect to y

$\int_{0}^{x} x y d y = x \int_{0}^{x} y d y = x [\frac{y ^{2}}{2}]_{0}^{x} = x (\frac{x ^{2}}{2} - 0) = \frac{x ^{3}}{2}$

Step 3: Integrate with Respect to x

$\int_{0}^{2} \frac{x ^{3}}{2} d x = \frac{1}{2} \int_{0}^{2} x^{3} d x = \frac{1}{2} [\frac{x ^{4}}{4}]_{0}^{2} = \frac{1}{2} (\frac{2 ^{4}}{4} - 0) = \frac{1}{2} \cdot \frac{16}{4} = \frac{8}{4} = 2$

Thus, the value of the double integral is 2.

Example 5: Evaluate the double integral $\iint_{R} sin (x + y) d A$ over the region R where $0 \leq x \leq \frac{π}{2}$ and

$0 \leq y \leq \frac{π}{2}$ .

Step 1: Set Up the Double Integral

$\iint_{R} sin (x + y) d A = \int_{0}^{\frac{π}{2}} (\int_{0}^{\frac{π}{2}} sin (x + y) d y) d x$

Step 2: Integrate with Respect to y

$\int_{0}^{\frac{π}{2}} sin (x + y) d y = [- cos (x + y)]_{0}^{\frac{π}{2}} = - cos (x + \frac{π}{2}) + cos (x) = sin (x) + cos (x)$

Step 3: Integrate with Respect to x

$\int_{0}^{\frac{π}{2}} (sin (x) + cos (x)) d x$

$= [- cos (x) + sin (x)]_{0}^{\frac{π}{2}}$

$= (- cos (\frac{π}{2}) + sin (\frac{π}{2})) - (- cos (0) + sin (0))$

=(0+1)-(-1+0)

=1+1=2

Thus, the value of the double integral is 2.

4.0Practice Questions on Double Integral

Question 1: Evaluate the double integral $\iint_{R} (3 x^{2} + 2 y) d A$ over the region R where $0 \leq x \leq 2$ and $0 \leq y \leq 1$ .

Question 2: Evaluate the double integral $\iint_{R} e^{x^{2} - y^{2}} d A$ over the region R where $0 \leq x \leq 1$ and $0 \leq y \leq 1$ .

Question 3: Evaluate the double integral $\iint_{R} (x + y) d A$ over the triangular region R with vertices at (0,0), (1,0), and (1,1).

Question 4: Evaluate the double integral $\iint_{R} cos (x y) d A$ over the region R where $0 \leq x \leq π$ and $0 \leq y \leq π$ .

Question 5: Evaluate the double integral $\iint_{R} \frac{1}{1 + x ^{2} + y ^{2}} d A$ over the region R where $0 \leq x \leq 1$ and $0 \leq y \leq 1$ .

Question 6: Evaluate the double integral $\iint_{R} x e^{x y} d A$ over the region R where $0 \leq x \leq 1$ and $0 \leq y \leq x$ .

Question 7: Evaluate the double integral $\iint_{R} sin (x + y) d A$ over the region R where $0 \leq x \leq \frac{π}{2}$ and $0 \leq y \leq \frac{π}{2}$ .

Question 8: Evaluate the double integral $\iint_{R} x^{2} y d A$ over the region R where $0 \leq x \leq 2$ and $0 \leq y \leq 3$ .

Question 9: Evaluate the double integral $\iint_{R} ln (x y) d A$ over the region R where $1 \leq x \leq 2$ and $1 \leq y \leq 2$ .

Question 10: Evaluate the double integral $\iint_{R} \frac{x ^{2} + y ^{2}}{1 + x ^{2} y ^{2}} d A$ over the region R where $0 \leq x \leq 1$ and $0 \leq y \leq 1$ .

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