Question

Evaluate the double integral.

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

Answer 1

`\iint_D 8y^2 \ dA = (88)/(3)`

Explanation:

The equation of the line through the point
`(x_o,y_o)` &
`(x_1,y_1)` can be represented by:

`y-y_o = m(x - x_o)`

Making m the subject;

`m = (y_1 - y_0)/(x_1-x_0)`

∴

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;

`m= (2-1)/(1-0)`

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;

`m = (1-2)/(4-1)`

`m = (-1)/(3)`

∴

`y-2 = -(1)/(3)(x-1)`

-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

∴

`\iint_D 8y^2 \ dA = \int^2_1 \int ^(-3y+7)_(y-1) \ 8y^2 \ dxdy`

`\iint_D 8y^2 \ dA =8 \int^2_1 \int ^(-3y+7)_(y-1) \ y^2 \ dxdy`

`\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^(-3y+7)_(y-1) \ dx \bigg) dy`

`\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^(-3y+7)_(y-1) \bigg ) \ dy`

`\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy`

`\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy`

`\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy`

`\iint_D 8y^2 \ dA =8 \bigg [( -4y^4)/(4)+(8y^3)/(3) \bigg ]^2_1`

`\iint_D 8y^2 \ dA =8 \bigg [ -y^4+(8y^3)/(3) \bigg ]^2_1`

`\iint_D 8y^2 \ dA =8 \bigg [ -2^4+(8(2)^3)/(3) + 1^4- (8* (1)^3)/(3)\bigg]`

`\iint_D 8y^2 \ dA =8 \bigg [ -16+(64)/(3) + 1- (8)/(3)\bigg]`

`\iint_D 8y^2 \ dA =8 \bigg [ -15+ (64-8)/(3)\bigg]`

`\iint_D 8y^2 \ dA =8 \bigg [ -15+ (56)/(3)\bigg]`

`\iint_D 8y^2 \ dA =8 \bigg [ (-45+56)/(3)\bigg]`

`\iint_D 8y^2 \ dA =8 \bigg [ (11)/(3)\bigg]`

`\iint_D 8y^2 \ dA = (88)/(3)`