Calculate the double integral: $\iint_R \frac{3xy^2}{x^2+1} \, dA, \quad R = \{(x,y) | 0 \leq x \leq 3, -2 \leq y \leq 2\}$

Neetesh Kumar | November 30, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 15.1 Question Number 14
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Step-by-step solution:

Step 1: Set up the iterated integral

The region $R$ is given by $0 \leq x \leq 3$ and $-2 \leq y \leq 2$, so the iterated integral becomes:

$\int_{-2}^2 \int_0^3 \frac{3xy^2}{x^2 + 1} \, dx \, dy$

Step 2: Evaluate the Inner Integral with Respect to $x$

We first focus on the inner integral:

$\int_0^3 \frac{3xy^2}{x^2 + 1} \, dx$

Since $y^2$ is a constant with respect to $x$, we can factor it out of the integral:

$y^2 \int_0^3 \frac{3x}{x^2 + 1} \, dx$

Now, we need to evaluate:

$\int_0^3 \frac{3x}{x^2 + 1} \, dx$

This is a standard integral, and we can use substitution. Let:

$u = x^2 + 1, \quad du = 2x \, dx$

Thus, the integral becomes:

$\frac{3}{2} \int_1^{10} \frac{du}{u}$

We used the fact that when $x = 0$, $u = 1$, and when $x = 3$, $u = 10$. Now, we integrate:

$\frac{3}{2} \ln(u) \Big|_1^{10} = \frac{3}{2} (\ln(10) - \ln(1))$

Since $\ln(1) = 0$, this simplifies to:

$\frac{3}{2} \ln(10)$

Thus, the inner integral becomes:

$y^2 \cdot \frac{3}{2} \ln(10)$

Step 3: Evaluate the Outer Integral with Respect to $y$

Now, substitute this result into the outer integral:

$\int_{-2}^2 y^2 \cdot \frac{3}{2} \ln(10) \, dy$

We can factor out the constant:

$\frac{3}{2} \ln(10) \int_{-2}^2 y^2 \, dy$

Now we need to evaluate:

$\int_{-2}^2 y^2 \, dy$

This is a straightforward integral. The integral of $y^2$ is:

$\int y^2 \, dy = \frac{y^3}{3}$

Evaluating from $y = -2$ to $y = 2$:

$\left[ \frac{y^3}{3} \right]_{-2}^2 = \frac{2^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \frac{-8}{3} = \frac{16}{3}$

Step 4: Combine the Results

Now, we multiply everything together:

$\frac{3}{2} \ln(10) \cdot \frac{16}{3} = 8 \ln(10)$

Final Answer:

The value of the double integral is:

$\boxed{8 \ln(10)}$

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