Evaluate the double integral. 4y2 dA, D is the triangular region with vertices (0, 1), (1, 2), (4, 1)

To answer this question, we need, first get the equations for the lines that enclosed the surface, and integrate according to the limits obtained from these equations give.

The solution is:

A = 44/3 square units  

Let´s call points:

P ( 0 , 1 )    Q ( 1 , 2 ) and R ( 4 , 1 )

The equation for the line between, P and R is:

y = 1

The equation for the line between, P and Q is:

Slope-intercept equation is  y = m×x + b

The slope   m₁ = ( 2 - 1 ) / ( 1 - 0 )    m₁ = 1

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

1 = 0 +b           b = 1

y = x + 1            ⇒    x = y - 1  

The equation  for the line between Q and R is:

m₂ = ( 1 - 2 ) / ( 4 - 1)    m₂ = - 1/3

y = ( -1/3)× x + b

when x = 1    y = 2

2 = ( - 1/3)×(1) + b

2 + 1/3 = b

b = 7/3

y = - (x/3) + 7/3           ⇒  x = 7 - 3×y

The double  integral becomes:

A = 4×∫∫ y² dx dy          ⇒   A = 4 ×∫₁² y²dy ∫dx  | (y - 1 ) y ( 7 - 3y)

A = 4×∫₁² y²dy  × x |  ( y - 1 ) y ( 7 - 3y)

A = 4 ×∫₁² y²dy  × [ 7 - 3×y - ( y - 1 )]

A = 4 ×∫₁² y²dy  × (8 - 4×y )      ⇒     A = 4 ×∫₁² (8×y² - 4×y³ ) dy

A = 4 × [ (8/3)×y³ - y⁴ | ₁²

A = 4 × [ 64/3 - 16 - (8/3) + 1 ]

A = 4  × ( 56/3 - 15 )

A = 4  × ( 56 - 45 /3)

A = 4 × 11/3

A = 44/3 square units

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44/3

Step-by-step explanation:

Let A be the line joining the vertices (0, 1) and (1,2) while B be the equation of the line joining (1, 2) and (4, 1).

The equation of a line joining points

The equation of line A joining (0, 1) and (1,2) is:

The equation of line D joining (1, 2) and (4,1) is:

Therefore the change in y is: 1 ≤ y ≤ 2, while change in x is: y-1 ≤ x ≤ -3y + 7. Hence the double integral is:

⊂⊃⊂⊃ω

step-by-step explanation:

i hope this : )