# Evaluate the double integral y^2 dA, D is the triangular region with vertices (0, 1), (1,2), (4,1)

This article aims to find the double integral of the triangular region with vertices. This article uses the concept of double integration. The definite integral of a positive function of one variable represents the area of the region between graph of the function and the $x-axis$. Similarly, the double integral of a positive function of two variables represents the volume of the region between the defined surface function (on the three-dimensional Cartesian plane, where $z = f(x, y)$ ) and the plane that contains its domain.

The points are:

$P (0,1) , Q(1,2) \: and \: R(4,1)$

The equation of line between $P$ and $R$ is given as:

$y = 1$

The equation of line between $P$ and $Q$ is given as:

Slope-intercept equation is given as:

$y = mx +c$

The slope is:

$m_{1} = \dfrac{2-1}{1-0} =1$

$m_{1} = 1$

and the line is passing over the point:

$x = 0\: , y = 1$

$1 = 0+ b$

$b = 1$

$y = x+1$

$x = y-1$

The equation for the line between $Q$ and $R$ is:

$m_{2} = \dfrac{(1-2)}{(4-1)} = -\dfrac{1}{3}$

$y = (-\dfrac{1}{3}) \times x +b$

$x =1 , y =2$

$2 = (-\dfrac{1}{3}) \times 1+ b$

$b = \dfrac{7}{3}$

$y = -(\dfrac{x}{3})+ \dfrac{7}{3}$

$x = 7 -3y$

The double integral becomes:

$A = \int \int y^{2} dx dy$

$A = \int y^{2} dy\int dx$

$A = \int y^{2} dy\int dx$

$A = \int_{1}^{2} y^{2} dy \times x |_{(y-1)}^{(7-3y)}$

$A = \int_{1}^{2} y^{2} dy \times (7-3y) – (y-1)$

$A = \int_{1}^{2} y^{2} dy \times (8-4y )$

$A = \int_{1}^{2} (8 y^{2} -4y^{3}) dy$

$= (\dfrac{8}{3} y^{3} – y^{4}) |_{1}^{2}$

$= \dfrac{56}{3} -15$

$A = \dfrac{11}{3}$

## Numerical Result

The solution is $A = \dfrac{11}{3}\: square\:units$.

## Example

Evaluate the double integral. $4 y^{2}\: dA$, $D$ is a triangular region with vertices $(0, 1), (1, 2), (4, 1)$.

Solution

The points are:

$P (0,1) , Q(1,2) \: and \: R(4,1)$

The equation of line between $P$ and $R$ is given as:

$y = 1$

The equation of line between $P$ and $Q$ is given as:

Slope-intercept equation is given as:

$y = mx +c$

The slope is:

$m_{1} = \dfrac{2-1}{1-0} =1$

$m_{1} = 1$

and the line is passing over the point:

$x = 0\: , y = 1$

$1 = 0+ b$

$b = 1$

$y = x+1$

$x = y-1$

The equation for the line between $Q$ and $R$ is:

$m_{2} = \dfrac{(1-2)}{(4-1)} = -\dfrac{1}{3}$

$y = (-\dfrac{1}{3}) \times x +b$

$x =1 , y =2$

$2 = (-\dfrac{1}{3}) \times 1+ b$

$b = \dfrac{7}{3}$

$y = -(\dfrac{x}{3})+ \dfrac{7}{3}$

$x = 7 -3y$

The double integral becomes:

$A = 4\int \int y^{2} dx dy$

$A = 4\int y^{2} dy\int dx$

$A = 4\int y^{2} dy\int dx$

$A = 4\int_{1}^{2} y^{2} dy \times x |_{(y-1)}^{(7-3y)}$

$A = 4\int_{1}^{2} y^{2} dy \times (7-3y) – (y-1)$

$A = 4\int_{1}^{2} y^{2} dy \times (8-4y )$

$A = 4\int_{1}^{2} (8 y^{2} -4y^{3}) dy$

$= 4(\dfrac{8}{3} y^{3} – y^{4}) |_{1}^{2}$

$=4(\dfrac{56}{3} -15)$

$A = 4(\dfrac{11}{3})$

$A = \dfrac{44}{3}$

The solution is $A = \dfrac{44}{3}\: square\:units$.