Evaluate the double integral $\iint_D 6xy \, dA$ where $D$ is the triangular region with vertices $(0,0)$ , $(1,2)$ , and $(0,3)$ .

Question :

Evaluate the double integral $\iint_d 6xy \, da$ where $d$ is the triangular region with vertices $(0,0)$ , $(1,2)$ , and $(0,3)$ .

$Evaluate the double integral \iint_d 6xy \, da where d is the triangular reg | Doubtlet.com$

Solution:

Neetesh Kumar | November 12, 2024

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This is the solution to Double Integral Question

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Step-by-step solution:

To evaluate the double integral $\iint_D 6xy \, dA$ , we need to set up the limits of integration for the region $D$ , which is defined by the triangular area with vertices $(0,0)$ , $(1,2)$ , and $(0,3)$ .

Step 1: Determine the Boundaries of the Region $D$

The vertices $(0,0)$ , $(1,2)$ , and $(0,3)$ form a triangular region in the $xy$ -plane. To set up the limits, we need to identify the equations of the lines that form the boundaries of this triangle.

Line from $(0,0)$ to $(1,2)$ :
The slope of this line is $\frac{2 - 0}{1 - 0} = 2$ , so the equation of this line is: $y = 2x$
Line from $(0,3)$ to $(1,2)$ :
The slope of this line is $\frac{2 - 3}{1 - 0} = -1$ , so the equation of this line is: $y = -x + 3$
Line $x = 0$ :
The Left vertical line forms the boundary at $x = 0$ .
Line $x = 0$ :
The Right vertical line forms the boundary at $x = 1$ .

Step 2: Set Up the Integral

The region $D$ can be described by the inequality $0 \leq x \leq 1$ , and for each $x$ , $y$ ranges from $y = 2x$ to $y = -x + 3$ .

Thus, we can set up the integral as follows: $\iint_D 6xy \, dA = \int_0^1 \int_{2x}^{-x + 3} 6xy \, dy \, dx$

Step 3: Integrate with Respect to $y$

First, we integrate the inner integral with respect to $y$ : $= \int_0^1 \int_{2x}^{-x + 3} 6xy \, dy \, dx$ $= \int_0^1 6x \left[ \frac{y^2}{2} \right]_{2x}^{-x + 3} \, dx$ $= \int_0^1 6x \left( \frac{(-x + 3)^2}{2} - \frac{(2x)^2}{2} \right) \, dx$

Step 4: Simplify the Expression

Expanding and simplifying the terms inside the integral:

Compute $(-x + 3)^2$ : $(-x + 3)^2 = x^2 - 6x + 9$
Compute $(2x)^2$ : $(2x)^2 = 4x^2$

Substitute these into the integral

$\int_0^1 6x \left( \frac{x^2 - 6x + 9}{2} - \frac{4x^2}{2} \right) \, dx$ $= \int_0^1 6x \left( \frac{x^2 - 6x + 9 - 4x^2}{2} \right) \, dx$ $= \int_0^1 6x \left( \frac{-3x^2 - 6x + 9}{2} \right) \, dx$ $= \int_0^1 3x (-3x^2 - 6x + 9) \, dx$ $= \int_0^1 (-9x^3 - 18x^2 + 27x) \, dx$

Step 5: Integrate with Respect to $x$

Now, integrate each term: $= \left[ -\frac{9x^4}{4} - 6x^3 + \frac{27x^2}{2} \right]_0^1$

Substitute $x = 1$ : $= -\frac{9}{4} - 6 + \frac{27}{2}$ $= -\frac{9}{4} - \frac{12}{2} + \frac{27}{2}$ $= -\frac{9}{4} + \frac{15}{2} = \frac{21}{4}$

Final Answer:

$\iint_D 6xy \, dA = \boxed{\frac{21}{4}}$

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Evaluate the double integral ∬D6xy dA\iint_D 6xy \, dA∬D​6xydA where DDD is the triangular region with vertices (0,0)(0,0)(0,0), (1,2)(1,2)(1,2), and (0,3)(0,3)(0,3).