Question

Evaluate the following double integral: xy dA D where the region D is the triangular region whose vertices are (0, 0), (0, 3), (3, 0).

Answer 1

Answer:

I= 84

Step-by-step explanation:

for

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{}_{} \int\limits^{}_D {x*y} \, dx*dy$

since D is the rectangle such that 0<x<3 , 0<y<3

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{3}_{0} \int\limits^{3}_{0} {x*y} \, dx*dy = \int\limits^{3}_{0} {x} \, dx\int\limits^{3}_{0} {y} \, dy = x^{2} /2*y^{2} /2 = (3^{2} /2 - 0^{2} /2)* (3^{2} /2 - 0^{2} /2) = 3^{4} /4 = 81/4$

Answer 2

Double Integral $\int\limits^{}_{} \int\limits^{}_D {xy} \, dA = \frac{81}{4} square units$

To answer the question, we need to know what double integrals are

What are double integrals?

These are integrals of two-dimensional surfaces.

Double Integral $\int\limits^{}_{} \int\limits^{}_D {xy} \, dA$

Given that region D is the triangular region whose vertices are (0, 0), (0, 3), (3, 0), we have that the interval of integration is 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3.

So, $\int\limits^{}_{} \int\limits^{}_D {xy} \, dA = \int\limits^3_0 \int\limits^3_0 {xy} \, dA \\\int\limits^{}_{} \int\limits^{}_D {xy} \, dA = \int\limits^3_0 \int\limits^3_0 {xy} \, dxdy$

$\int\limits^3_0 \int\limits^3_0 {xy} \, dxdy = \int\limits^3_0 {x} \,dx \int\limits^3_0 {y} \, dy\\= \int\limits^3_0 {x} \,dx [\frac{y^{2} }{2}] _{0}^{3} \\= \int\limits^3_0 {x} \,dx [\frac{3^{2} }{2} - \frac{0}{2} ]\\= \int\limits^3_0 {x} \,dx [\frac{9}{2} - 0 ]\\= \frac{9}{2}\int\limits^3_0 {x} \,dx \\= \frac{9}{2}[\frac{x^{2} }{2}] _{0}^{3} \\= \frac{9}{2}[\frac{3^{2} }{2} - \frac{0}{2} ]\\= \frac{9}{2}[\frac{9}{2} - 0]\\= \frac{9}{2}[\frac{9}{2} ]\\= \frac{81}{4} square units$

So, double Integral $\int\limits^{}_{} \int\limits^{}_D {xy} \, dA = \frac{81}{4} square units$

Learn more about double integrals here:

https://brainly.com/question/19053586