Multiple Integrals : Double Integral

Evaluation of Double Integral

Consider a function f(x , y) of two variables x and y defined in a region A of X-Y plane . Double Integral of a function f(x , y) over a region A be evaluated by two successive integrations .

If A is described as f1 (x) ≤ y ≤ f2 [ y1 ≤ y ≤ y2 ] and a ≤ x ≤ b , Then

$\displaystyle \int \int_A f(x , y)dA = \int_{a}^{b} \int_{y_1}^{y_2} f(x , y) dx dy$

Example : Evaluate $\int \int_R x y dx dy $ ; Where R is the quadrant of circle x2 + y2 = a2 where x ≥ 0 and y ≥ 0

Solution : The Region of Integration is first quadrant of the circle .

x2 + y2 = a2 ; $y = \sqrt{a^2 – x^2}$

First Integrate with respect to y and then with respect to x . The limit of x are 0 to a and for y , 0 to √(a2 -x2)

$\displaystyle I = \int_{0}^{a} x dx \int_{0}^{\sqrt{a^2 – x^2}}y dy $

$\displaystyle I = \int_{0}^{a} x dx [\frac{y^2}{2}]_{0}^{\sqrt{a^2 – x^2}} $

$\displaystyle I = \frac{1}{2} \int_{0}^{a} x (a^2 – x^2) dx $

$\displaystyle I = \frac{1}{2} [\frac{a^2 x^2}{2} – \frac{x^4}{4}]_{0}^{a} = \frac{a^4}{4} $

Example: Evaluate the double integral $\int \int_R x y dx dy $ ; where R is the region bounded by the X-axis , the line y = 2 x and the parabola x2 = 4 a y

Solution: Solving y = 2x and x2 = 4 a y

x2 = 4 a (2x)

x2 – 8 a x = 0

x (x – 8 a ) = 0

x = 0 , 8 a

Engg. Math

$\displaystyle \int \int_R x y dx dy $

$\displaystyle = \int_{0}^{8a} x dx \int_{x^2/4a}^{2x} y dy $

$\displaystyle = \int_{0}^{8a} x dx [\frac{y^2}{2}]_{x^2/4a}^{2x} $

$\displaystyle = \int_{0}^{8a} x dx [2x^2 – \frac{x^4}{32a^2}] $

$\displaystyle = \int_{0}^{8a} [2x^3 – \frac{x^5}{32a^2}] dx $

$\displaystyle = [2\frac{x^4}{4} – \frac{x^6}{6 \times 32a^2}]_{0}^{8a} $

$\displaystyle = [\frac{x^4}{2} – \frac{x^6}{192 a^2}]_{0}^{8a} $

$\displaystyle = \frac{2048}{3} a^4 $

Example: Find the area between the parabolas y2 = 4 a x and x2 = 4 a y

Solution: On solving the two equations we get the point of intersection (4a , 4a)

Engg. Math

Now divide the area into horizontal strips & integrating the area of strip under limits gives Required Area .

The required area $\displaystyle = \int_{0}^{4a} dy \int_{y^2/4a}^{\sqrt{4ay}} dx $
(Since x varies from y2/4a to √(4ay))

$\displaystyle = \int_{0}^{4a} dy [x]_{y^2/4a}^{\sqrt{4ay}} $

$\displaystyle = \int_{0}^{4a} dy [\sqrt{4ay} – \frac{y^2}{4a}]_{y^2/4a}^{\sqrt{4ay}} $

$\displaystyle = [\sqrt{4a}\frac{y^{3/2}}{3/2} – \frac{y^3}{12 a}]_{0}^{4a} $

$\displaystyle = [\frac{4 \sqrt{a}}{3}(4a)^{3/2} – \frac{(4a)^3}{12 a}] $

$\displaystyle = \frac{32}{3} a^2 – \frac{16}{3}a^2 = \frac{16}{3}a^2$

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