Question

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

Answer 1

Answer:

$\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}$

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:

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

y - 1 = x

The line equtaion for (1,2),(4,1) is:

$y-2 = \dfrac{1-2}{4-1}(x-1) \\ \\ y-2 = -\dfrac{1}{3}(x-1)$

-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:

$\iint _D y^2 dA= \int ^2_1 \int ^{-3y +7}_{y-1} \ 2y ^2 \ dx \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[ 2xy ^2 \bigg]^{-3y+7}_{y-1} \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[2(-3y+7)y^2-2(y-1)y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[-6y^3 +14y^2 -2y^3 +2y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[-8y^3 +16y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \bigg[-8(\dfrac{y^4}{4}) +16(\dfrac{y^3}{3})\bigg ] ^2_1$

$\iint _D y^2 dA= \bigg[-8(\dfrac{16}{4}-\dfrac{1}{4}) +16(\dfrac{8}{3}-\dfrac{1}{3})\bigg ]$

$\iint _D y^2 dA= \bigg[-8(\dfrac{15}{4}) +16(\dfrac{7}{3})\bigg ]$

$\iint _D y^2 dA= -30 + \dfrac{112}{3}$

$\iint _D y^2 dA= \dfrac{-90+112}{3}$

$\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}$