To evaluate the given triple integral:

$\int_0^{y}\int_0^{\sqrt3z}\frac{z}{x^2+z^2} \, dx \, dz \, dy$

Step-by-Step Evaluation

1. Inner Integral with Respect to $x$:

The integrand is $\frac{z}{x^2+z^2}$. Here, $z$ is treated as a constant with respect to $x$.

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

Let’s perform a substitution: $u = x^2 + z^2 \quad \Rightarrow \quad du = 2x \, dx$ For $x = 0$, $u = z^2$. For $x = \sqrt{3}z$, $u = (\sqrt{3}z)^2 + z^2 = 4z^2$.

Using this substitution, we rewrite: $\int_0^{\sqrt{3}z} \frac{z}{x^2+z^2} \, dx = \int_{z^2}^{4z^2} \frac{z}{u} \cdot \frac{1}{2z} \, du = \frac{1}{2} \int_{z^2}^{4z^2} \frac{1}{u} \, du$

The integral of $\frac{1}{u}$ is $\ln|u|$: $\frac{1}{2} \int_{z^2}^{4z^2} \frac{1}{u} \, du = \frac{1}{2} \big[ \ln(4z^2) - \ln(z^2) \big] = \frac{1}{2} \ln\left(\frac{4z^2}{z^2}\right) = \frac{1}{2} \ln(4) = \ln(2)$

So, the result of the inner integral is: $\int_0^{\sqrt{3}z} \frac{z}{x^2+z^2} \, dx = \ln(2)$

2. Middle Integral with Respect to $z$:

Next, substitute this result into the middle integral: $\int_0^y \ln(2) \, dz$

Since $\ln(2)$ is a constant: $\int_0^y \ln(2) \, dz = \ln(2) \int_0^y 1 \, dz = \ln(2) \cdot y$

3. Outer Integral with Respect to $y$:

Finally, substitute this into the outer integral: $\int_0^{y} y \ln(2) \, dy$

This is: $\ln(2) \int_0^y y \, dy = \ln(2) \cdot \frac{y^2}{2} \Big|_0^1 = \ln(2) \cdot \frac{1^2}{2} = \frac{\ln(2)}{2}$

Final Answer:

$\int_0^{y} \int_0^{\sqrt{3}z} \frac{z}{x^2+z^2} \, dx \, dz \, dy = \frac{\ln(2)}{2}$

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Do you want detailed clarification or further steps explained?

Related Questions:

1. What is the role of substitution in evaluating integrals?
2. How does the logarithmic function arise in integrals involving $\frac{1}{u}$?
3. How does symmetry simplify multi-variable integration problems?
4. What is the geometric interpretation of the integrand in this problem?
5. Can this integral be solved using polar coordinates?

Tip: Always ensure that variable limits and substitutions are consistent with the original integral setup.