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Sever21 [200]
3 years ago
9

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

Mathematics
1 answer:
zubka84 [21]3 years ago
4 0

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}}

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