Answer :
Answer:
[tex]\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}[/tex]
Step-by-step explanation:
From the image attached below;
We need to calculate the limits of x and y to find the double integral
We will notice that y varies from 1 to 2
The line equation for (0,1),(1,2) is:
[tex]y-1 = \dfrac{2-1}{1-0}(x-0)[/tex]
y - 1 = x
The line equtaion for (1,2),(4,1) is:
[tex]y-2 = \dfrac{1-2}{4-1}(x-1) \\ \\ y-2 = -\dfrac{1}{3}(x-1)[/tex]
-3(y-2) = (x -1)
-3y + 6 = x - 1
-x = 3y - 6 - 1
-x = 3y - 7
x = -3y + 7
This implies that x varies from y - 1 to -3y + 7
Now, the region D = {(x,y) | 1 ≤ y ≤ 2, y - 1 ≤ x ≤ -3y + 7}
The double integral can now be calculated as:
[tex]\iint _D y^2 dA= \int ^2_1 \int ^{-3y +7}_{y-1} \ 2y ^2 \ dx \ dy[/tex]
[tex]\iint _D y^2 dA= \int ^2_1 \bigg[ 2xy ^2 \bigg]^{-3y+7}_{y-1} \ dy[/tex]
[tex]\iint _D y^2 dA= \int ^2_1 \bigg[2(-3y+7)y^2-2(y-1)y^2 \bigg ] \ dy[/tex]
[tex]\iint _D y^2 dA= \int ^2_1 \bigg[-6y^3 +14y^2 -2y^3 +2y^2 \bigg ] \ dy[/tex]
[tex]\iint _D y^2 dA= \int ^2_1 \bigg[-8y^3 +16y^2 \bigg ] \ dy[/tex]
[tex]\iint _D y^2 dA= \bigg[-8(\dfrac{y^4}{4}) +16(\dfrac{y^3}{3})\bigg ] ^2_1[/tex]
[tex]\iint _D y^2 dA= \bigg[-8(\dfrac{16}{4}-\dfrac{1}{4}) +16(\dfrac{8}{3}-\dfrac{1}{3})\bigg ][/tex]
[tex]\iint _D y^2 dA= \bigg[-8(\dfrac{15}{4}) +16(\dfrac{7}{3})\bigg ][/tex]
[tex]\iint _D y^2 dA= -30 + \dfrac{112}{3}[/tex]
[tex]\iint _D y^2 dA= \dfrac{-90+112}{3}[/tex]
[tex]\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}[/tex]
To answer this question, we need, first get the equations for the lines that enclosed the surface, and integrate according to the limits these equations give.
The solution is:
A = 22/3 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 = 2×∫∫ y² dx dy ⇒ A = 2 ×∫₁² y²dy ∫dx | (y - 1 ) y ( 7 - 3y)
A = 2 ×∫₁² y²dy × x | ( y - 1 ) y ( 7 - 3y)
A = 2 ×∫₁² y²dy × [ 7 - 3×y - ( y - 1 )]
A = 2 ×∫₁² y²dy × (8 - 4×y ) ⇒ A = 2 ×∫₁² (8×y² - 4×y³ ) dy
A = 2 × [ (8/3)×y³ - y⁴ | ₁²
A = 2 × [ 64/3 - 16 - (8/3) + 1 ]
A = 2 × ( 56/3 - 15 )
A = 2 × ( 56 - 45 /3)
A = 2 × 11/3
A = 22/3 square units
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