91Ó°ÊÓ

To convert the given integral to cylindrical coordinates, we will replace the coordinates \((x, y, z)\) with \((r, \theta, z)\), where \(r^2 = x^2 + y^2\) and the Jacobian is given by \(J = r\). Let's find the bounds for \(r\), \(\theta\), and \(z\): - For \(x\), we have \(-3 \le x \le 3\). Since \(x = r\cos \theta\), we can rewrite this as \(0 \le r\cos \theta \le 3\). - For \(y\), we have \(0 \le y \le \sqrt{9 - x^2}\). Since \(y = r\sin \theta\), we can rewrite this as \(0 \le r\sin \theta \le \sqrt{9 - r^2\cos^2\theta}\). However, in cylindrical coordinates, \(y\) has limits of \(0 \le y \le 2\pi\). - For \(z\), we have \(0 \le z \le 2\). So the transformed integral becomes: $$\int \int \int \frac{r}{1+r^2} d z d r d \theta$$ Now that we've found the bounds let's proceed to evaluate the integrals.

We first evaluate the inner integral with respect to \(z\): $$\int_{0}^{2} \frac{1}{1+r^2} dz$$ $$= \left[\frac{z}{1+r^2}\right]_{0}^{2} = \frac{2}{1+r^2}$$

Now we evaluate the middle integral with respect to \(r\): $$\int_{0}^{\sqrt{9-x^2}} \frac{2}{1+r^2} r dr$$ Make a substitution: \(u = 1 + r^2, du = 2r dr\): $$\int_{1}^{10} \frac{r du}{u} = \int_{1}^{10} \frac{du}{u}$$ $$= \left[\ln|u|\right]_{1}^{10} = \ln(\frac{10}{1}) = \ln(10)$$ Keep in mind that in this phase, we just find the integral, but we haven't substituted back the variable yet.

Finally, we evaluate the outer integral with respect to \(\theta\): $$\int_{-3}^{3} \ln(10) d\theta = \ln(10)\left[\theta\right]_{-3}^{3} = 6\ln(10)$$ So, the value of the given triple integral in cylindrical coordinates is: $$6\ln(10)$$