Mathematics, 28.10.2020 16:40 kailibug2287

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

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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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Answer from: KallMeh

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 $(x_1,y_1) \ and\ (x_2,y_2)\ is:\\\\$

$y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)$

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

$y-1=\frac{2-1}{1-0} (x-0)\\\\y-1=x\\\\x=y-1$

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

$y-2=\frac{1-2}{4-1} (x-1)\\\\y-2=-\frac{1}{3} (x-1)\\\\3y-6=-x+1\\\\x=-3y+7$

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

$\int\limits^2_1\int\limits^{-3y+7}_{y-1} {4y^2} \, dx dy\\\\=4\int\limits^2_1\int\limits^{-3y+7}_{y-1} {y^2} \, dx dy\\\\=4\int\limits^2_1y^2dy[x]^{-3y+7}_{y-1} \\\\=4\int\limits^2_1y^2dy(-3y+7-(y-1))\\\\=4\int\limits^2_1y^2dy(-4y+8)\\\\=4\int\limits^2_1(-4y^3+8y^2)dy\\\\=4[-y^4+\frac{8}{3}y^3 ]^2_1\\\\=4(\frac{11}{3} )\\\\=\frac{44}{3 }$