Answer:
[tex]\iint_D 8y^2 \ dA = \dfrac{88}{3}[/tex]
Step-by-step explanation:
The equation of the line through the point [tex](x_o,y_o)[/tex] & [tex](x_1,y_1)[/tex] can be represented by:
[tex]y-y_o = m(x - x_o)[/tex]
Making m the subject;
[tex]m = \dfrac{y_1 - y_0}{x_1-x_0}[/tex]
∴
we need to carry out the equation of the line through (0,1) and (1,2)
i.e
y - 1 = m(x - 0)
y - 1 = mx
where;
[tex]m= \dfrac{2-1}{1-0}[/tex]
m = 1
Thus;
y - 1 = (1)x
y - 1 = x ---- (1)
The equation of the line through (1,2) & (4,1) is:
y -2 = m (x - 1)
where;
[tex]m = \dfrac{1-2}{4-1}[/tex]
[tex]m = \dfrac{-1}{3}[/tex]
∴
[tex]y-2 = -\dfrac{1}{3}(x-1)[/tex]
-3(y-2) = x - 1
-3y + 6 = x - 1
x = -3y + 7
Thus: for equation of two lines
x = y - 1
x = -3y + 7
i.e.
y - 1 = -3y + 7
y + 3y = 1 + 7
4y = 8
y = 2
Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7
∴
[tex]\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy[/tex]
[tex]\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy[/tex]
[tex]\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy[/tex]
[tex]\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy[/tex]
[tex]\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy[/tex]
[tex]\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy[/tex]
[tex]\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy[/tex]
[tex]\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1[/tex]
[tex]\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1[/tex]
[tex]\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg][/tex]
[tex]\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg][/tex]
[tex]\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg][/tex]
[tex]\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg][/tex]
[tex]\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg][/tex]
[tex]\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg][/tex]
[tex]\iint_D 8y^2 \ dA = \dfrac{88}{3}[/tex]
