Question

evaluate the double integral. 7y2 da, d is the triangular region with vertices (0, 1), (1, 2), (4, 1) d

Answer 1

The value of double integral over the triangular region is $\frac{77}{3}$ square units.

Given,

the vertices of triangular region are A(0, 1), B(1, 2), C(4, 1).

the double integral function is $\int\limits\int\limits 7y^2 dA$

The equation for line between A and C is:

y = 1

The equation for line between A and B is:

General equation is  $y = mx + b$

The slope,

$m_1 = \frac{ 2 - 1}{ 1 - 0 }\\\\ m_1 = 1$

and the line passes over the point  x = 0  y = 1; then

$1 = 0 +b \\\\b = 1$

equation becomes,

$y = x + 1\\\\ x = y - 1$

The equation  for line between B and C is:

$m_2 = \frac{ 1 - 2 }{ 4 - 1}\\\\m_2 = \frac{- 1}{3}$

$y = \frac{-1}{3}x + b$

when (x, y) = (1, 2)

$2 = \frac{- 1}{3}(1) + b\\\\2 + \frac{1}{3} = b\\\\b =\frac{7}{3}$

Equation becomes,

$y = - \frac{x}{3} + \frac{7}{3}\\\\x = 7 - 3y$

The double integral becomes,

$=\int\limits\int\limits 7y^2 dA\\\\\int\limits^2_1\int\limits^{7-3y}_{y-1} 7y^2 dxdy\\\\\int\limits^{2}_{y=1} 7 y^2dy\int\limits^{7-3y}_{x=y-1}dx\\\\\int\limits^{2}_{y=1} 7 y^2dy[x]|\limits^{7-3y}_{x=y-1}\\\\\int\limits^{2}_{y=1} 7 y^2dy[7-3y-(y-1)]\\\\=7[\frac{y^3}{3}][8-4y]|\limits^2_1\\\\=7[\frac{11}{3}]\\\\=\frac{77}{3}$

Thus the value of double integral over the triangular region is $\frac{77}{3}$ square units.

To learn more about integration refer here

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