The integral shown in the image is:

$\int_0^{\pi} \int_1^2 \left( 3r \cos \theta + r^2 \sin^2 \theta \right) \, dr \, d\theta$

To evaluate this double integral, let's proceed with the integration in two steps:

Step 1: Inner Integral with respect to $r$

We have: $\int_1^2 \left( 3r \cos \theta + r^2 \sin^2 \theta \right) \, dr$

Breaking it down: $= \int_1^2 3r \cos \theta \, dr + \int_1^2 r^2 \sin^2 \theta \, dr$

1. Evaluating $\int_1^2 3r \cos \theta \, dr$:

   $= 3 \cos \theta \int_1^2 r \, dr = 3 \cos \theta \left[ \frac{r^2}{2} \right]_1^2 = 3 \cos \theta \cdot \frac{4 - 1}{2} = \frac{9}{2} \cos \theta$

2. Evaluating $\int_1^2 r^2 \sin^2 \theta \, dr$:

   $= \sin^2 \theta \int_1^2 r^2 \, dr = \sin^2 \theta \left[ \frac{r^3}{3} \right]_1^2 = \sin^2 \theta \cdot \frac{8 - 1}{3} = \frac{7}{3} \sin^2 \theta$

So, the result of the inner integral is: $\frac{9}{2} \cos \theta + \frac{7}{3} \sin^2 \theta$

Step 2: Outer Integral with respect to $\theta$

Now we integrate this result with respect to $\theta$ from 0 to $\pi$: $\int_0^{\pi} \left( \frac{9}{2} \cos \theta + \frac{7}{3} \sin^2 \theta \right) d\theta$

We can split this into two integrals: $= \frac{9}{2} \int_0^{\pi} \cos \theta \, d\theta + \frac{7}{3} \int_0^{\pi} \sin^2 \theta \, d\theta$

1. Evaluating $\int_0^{\pi} \cos \theta \, d\theta$:

   Since $\cos \theta$ is an odd function over the symmetric interval $[0, \pi]$, this integral evaluates to zero: $\int_0^{\pi} \cos \theta \, d\theta = 0$

2. Evaluating $\int_0^{\pi} \sin^2 \theta \, d\theta$:

   Using the identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$: $\int_0^{\pi} \sin^2 \theta \, d\theta = \int_0^{\pi} \frac{1 - \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \int_0^{\pi} 1 \, d\theta - \frac{1}{2} \int_0^{\pi} \cos(2\theta) \, d\theta$

   The first part is: $\frac{1}{2} \int_0^{\pi} 1 \, d\theta = \frac{1}{2} \cdot \pi = \frac{\pi}{2}$

   The second part is zero since $\cos(2\theta)$ over $[0, \pi]$ integrates to zero: $\frac{1}{2} \int_0^{\pi} \cos(2\theta) \, d\theta = 0$

   So, $\int_0^{\pi} \sin^2 \theta \, d\theta = \frac{\pi}{2}$

Thus, the result of the outer integral is: $\frac{9}{2} \cdot 0 + \frac{7}{3} \cdot \frac{\pi}{2} = \frac{7\pi}{6}$

Final Answer:

$\int_0^{\pi} \int_1^2 \left( 3r \cos \theta + r^2 \sin^2 \theta \right) \, dr \, d\theta = \frac{7\pi}{6}$

Would you like further explanation on any part of this solution?

Related Questions:

1. How would the result change if the limits of $r$ were different?
2. What if we had $\sin \theta$ instead of $\sin^2 \theta$ in the integrand?
3. How do we interpret this result in a physical context?
4. Can this integral be evaluated using polar coordinates in a different way?
5. What if the outer integral was from $0$ to $2\pi$ instead?

Tip:

For integrals involving trigonometric functions squared, using identities like $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$ can simplify the calculation.