Let's solve the given double integral step by step: \[ \int_0^{2\pi} \int_0^a \frac{r}{\sqrt{a^2 - r^2}} \, dr \, dt \] ### **Step 1: Separate the Integrals** Notice that the limits for \( t \) and \( r \) are independent, so we can separate the integrals: \[ \int_0^{2\pi} dt \int_0^a \frac{r}{\sqrt{a^2 - r^2}} \, dr \] ### **Step 2: Solve the Inner Integral** Let's focus on the inner integral: \[ I = \int_0^a \frac{r}{\sqrt{a^2 - r^2}} \, dr \] Let’s use substitution. Let \( u = a^2 - r^2 \implies du = -2r dr \implies -\frac{1}{2} du = r dr \). - When \( r = 0 \), \( u = a^2 \) - When \( r = a \), \( u = 0 \) So the integral becomes: \[ I = \int_{r=0}^{r=a} \frac{r}{\sqrt{a^2 - r^2}} dr = \int_{u=a^2}^{u=0} \frac{1}{\sqrt{u}} \left(-\frac{1}{2} du \right) \] \[ = -\frac{1}{2} \int_{u=a^2}^{u=0} u^{-1/2} du \] Switch the limits and remove the negative sign: \[ = \frac{1}{2} \int_{u=0}^{u=a^2} u^{-1/2} du \] \[ = \frac{1}{2} \left[ 2 u^{1/2} \right]_0^{a^2} \] \[ = \left[ u^{1/2} \right]_0^{a^2} \] \[ = (a^2)^{1/2} - (0)^{1/2} = a - 0 = a \] ### **Step 3: Solve the Outer Integral** Now substitute back: \[ \int_0^{2\pi} a \, dt = a \int_0^{2\pi} dt = a [ t ]_0^{2\pi} = a (2\pi - 0) = 2\pi a \] --- ## **Final Answer** \[ \boxed{2\pi a} \] This is the value of the given double integral.

Let's carefully evaluate the given double integral step by step, providing detailed calculations and explanations. --- ### **Given integral:** \[ \iint_{D} \frac{r}{\sqrt{a^2 - r^2}} \, dr \, dt \] with bounds: - \( 0 < t < 2\pi \) - \( 0 < r < a \) --- ### **Step 1: Separate the integral** Since the limits for \( t \) and \( r \) are independent, the double integral can be written as a product of two integrals: \[ \left( \int_{t=0}^{2\pi} dt \right) \left( \int_{r=0}^{a} \frac{r}{\sqrt{a^2 - r^2}} dr \right) \] Thus, \[ \boxed{ \text{Integral} = \left( \int_0^{2\pi} dt \right) \times \left( \int_0^a \frac{r}{\sqrt{a^2 - r^2}} dr \right) } \] --- ### **Step 2: Evaluate the \( t \)-integral** \[ \int_0^{2\pi} dt = 2\pi \] --- ### **Step 3: Focus on the \( r \)-integral** \[ I = \int_0^a \frac{r}{\sqrt{a^2 - r^2}} dr \] This is a standard integral. To evaluate it, we use substitution: **Substitution:** \[ u = a^2 - r^2 \implies du = -2r dr \] which implies: \[ r dr = -\frac{1}{2} du \] **Change of limits:** - When \( r = 0 \): \[ u = a^2 - 0 = a^2 \] - When \( r = a \): \[ u = a^2 - a^2 = 0 \] **Rewrite the integral:** \[ I = \int_{r=0}^{a} \frac{r}{\sqrt{a^2 - r^2}} dr = \int_{u=a^2}^{0} \frac{1}{\sqrt{u}} \left( -\frac{1}{2} du \right) \] which simplifies to: \[ I = -\frac{1}{2} \int_{a^2}^0 u^{-1/2} du \] Switch the limits to get rid of the negative sign: \[ I = \frac{1}{2} \int_0^{a^2} u^{-1/2} du \] --- ### **Step 4: Integrate** Recall that: \[ \int u^{n} du = \frac{u^{n+1}}{n+1} + C \] for \( n \neq -1 \). Here, \( n = -\frac{1}{2} \), so: \[ \int u^{-1/2} du = 2 u^{1/2} + C \] Applying the limits: \[ I = \frac{1}{2} \times 2 u^{1/2} \bigg|_{0}^{a^2} = u^{1/2} \bigg|_{0}^{a^2} \] Evaluate at the limits: \[ I = (a^2)^{1/2} - (0)^{1/2} = a - 0 = a \] --- ### **Step 5: Combine the results** The original integral becomes: \[ \boxed{ \text{Integral} = (2\pi) \times a = 2 \pi a } \] --- ## **Final answer:** \[ \boxed{ \boxed{2\pi a} } \] This is the value of the given double integral. --- **Summary:** - The integral separates into a product because of independent bounds. - The inner integral reduces via substitution to a standard form. - After evaluation, the total integral evaluates to \( 2 \pi a \). Let me know if you'd like further clarification!