Question

Evaluate the integral xy dA, where R is the triangle x + y < 6,x > 0, y = 0. Enter the exact answer as improper fraction if necessary. be xy dA =

Answer 1

To evaluate the integral xy dA, where R is the triangle x + y < 6, x > 0, y = 0, we can set up the integral using the given limits and the equation xy. We can then use the power rule of integration to evaluate the integral and find the final answer.

Step-by-step explanation:

To evaluate the integral xy dA, where R is the triangle x + y < 6, x > 0, y = 0, we can set up the integral using the given limits and the equation xy. Since the triangle is defined by x + y < 6, we can express y in terms of x as y = 6 - x. The integral becomes ∫(x(6-x)) dA, which simplifies to ∫(6x - x^2) dx.

To evaluate this integral, we can use the power rule of integration. The integral of 6x is 3x^2, and the integral of x^2 is (1/3)x^3. Integrating these terms separately and evaluating them from 0 to 6 gives us the final answer.

Answer 2

The value of
`\( \iint_(R) x y \, dA \)` over the given region R is 54.

Given the region R defined by the inequalities
`\(x + y \leq 6\)`,
`\(x \geq 0\)`, and
`\(y \geq 0\)`, it forms a triangle in the first quadrant of the xy-plane.

The integral to evaluate is:

`\[ \iint_(R) x y \, dA \]`

To integrate over this region, we'll express the integral in terms of x and y. The limits of integration for x and y are based on the bounds of the triangle.

The bounds for x are from 0 to 6 (as
`\(x + y \leq 6\) gives \(x = 6 - y\)`).

The bounds for y are from 0 to 6 (as
`\(x + y \leq 6\) gives \(y = 6 - x\)`).

So, the integral becomes:

`\[ \iint_(R) x y \, dA = \int_(0)^(6) \int_(0)^(6-x) xy \, dy \, dx \]`

Let's solve it step by step:

`\[ \int_(0)^(6) \int_(0)^(6-x) xy \, dy \, dx \]`

`\[ = \int_(0)^(6) \left[x \cdot (y^2)/(2)\right]_(0)^(6-x) \, dx \]`

`\[ = \int_(0)^(6) x \cdot ((6-x)^2)/(2) \, dx \]`

`\[ = \int_(0)^(6) x \cdot (36 - 12x + x^2)/(2) \, dx \]`

`\[ = \int_(0)^(6) \left(18x - 6x^2 + (x^3)/(2)\right) \, dx \]`

`\[ = \left[(18x^2)/(2) - (6x^3)/(3) + (x^4)/(8)\right]_(0)^(6) \]`

`\[ = \left[9x^2 - 2x^3 + (x^4)/(8)\right]_(0)^(6) \]`

`\[ = 9(6)^2 - 2(6)^3 + ((6)^4)/(8) - 0 \]`

`\[ = 9(36) - 2(216) + 162 \]`

`\[ = 324 - 432 + 162 \]`

`\[ = 486 - 432 \]`

`\[ = 54 \]`

Therefore, after evaluating the integral, the value of

`\[ \iint_(R) x y \, dA = 54 \]`

Question:

Evaluate the integral
`\( \iint_(R) x y d A \)`, where R is the triangle
`\( x+y \leq 6, x \geq 0, y \geq 0 \)`. Enter the exact answer as improper fraction if necessary.
`\[\iint_(R) x y d A=\]`