Question

Find the double integral over D of 3xy^2 dA, where D is the region bounded by the points (0,0), (3,1), and (0,3).

Answer 1

Now you can easily solve it.

I hope you will get it.

Answer 2

Answer:

$I=\dfrac{243}{10}$

Step-by-step explanation:

The region bounded by the points (0,0), (3,1), and (0,3).

First draw the region and find the equation of each path.

Please find the attachment for region.

For the region, the limit of x and y

$0\leq x\leq 3$

$\dfrac{1}{3}x\leq y \leq -\dfrac{2}{3}x+3$

$dA=dydx$

Substitute into integral, $I=\iint 3xy^2dA$

$I=\int_0^3\int_{\frac{1}{3}x}^{-\frac{2}{3}x+3}3xy^2dydx$

Now integrate it

$I=\int_0^3xy^3|_{\frac{1}{3}x}^{-\frac{2}{3}x+3}3xy^2dx$

Apply limit upper limit - lower limit

$I=\int_0^3(-\dfrac{x^4}{3}+4x^3-18x^2+27x)dx$

$I=(-\dfrac{x^5}{15}+x^4-6x^3+27x^2/2)_0^3$

$I=-\dfrac{81}{5}+81-162+\dfrac{243}{2}$

$I=\dfrac{243}{10}$