Evaluate the double integral.
∫∫8y2dA, D is the triangular region with vertices (0, 1), (1, 2), (4, 1)

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]

You can use the interactive graph below to find the answer.
\begin{cases} y=x+1 \\\\ y=2x-5 \end{cases}
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​

y=x+1
y=2x−5
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Choose 1 answer:
Choose 1 answer:

(Choice A)
A
Exactly one solution

(Choice B)
B
No solutions

(Choice C)
C
Infinitely many solutions

[tex]2 log_{2}(x) + \frac{1}{2} log_{2}(x - 1) - log_{2}( x + 3) [/tex]
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