Evaluate Double Integral Using Polar Coordinates

Math Problem Statement

Determine the double integral $ \int_0^1 \int_0^{\sqrt{a^2 - y^2}} (x^2 + y^2) \ dy \ dx $ by changing into polar coordinates.

The problem involves evaluating the given double integral by converting it to polar coordinates.

The given integral is:

$\int_0^1 \int_0^{\sqrt{a^2 - y^2}} \left( x^2 + y^2 \right) \, dy \, dx$

Here, the integration bounds and the integrand suggest a circle (due to the presence of $a^2 - y^2$ ).

In polar coordinates:

The region of integration corresponds to a quarter-circle of radius $a$ in the first quadrant:

The integrand $x^2 + y^2$ becomes $r^2$ , and the differential becomes $r \, dr \, d\theta$ . The integral is transformed as:

\quad \to \quad \int_0^{\frac{\pi}{2}} \int_0^a r^2 \cdot r \, dr \, d\theta$$ Simplify the integrand: $$\int_0^{\frac{\pi}{2}} \int_0^a r^3 \, dr \, d\theta$$ --- #### 4. **Evaluate the Integral** 1. **Inner Integral ($$r$$)**: $$\int_0^a r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^a = \frac{a^4}{4}$$ 2. **Outer Integral ($$\theta$$)**: $$\int_0^{\frac{\pi}{2}} \frac{a^4}{4} \, d\theta = \frac{a^4}{4} \cdot \int_0^{\frac{\pi}{2}} 1 \, d\theta$$ $$\int_0^{\frac{\pi}{2}} 1 \, d\theta = \theta \Big|_0^{\frac{\pi}{2}} = \frac{\pi}{2}$$ Combine results: $$\frac{a^4}{4} \cdot \frac{\pi}{2} = \frac{\pi a^4}{8}$$ --- #### 5. **Final Answer** The value of the double integral is: $$\boxed{\frac{\pi a^4}{8}}$$ --- Do you need further clarification or a breakdown of any step? Let me know! --- ### 5 Related Questions: 1. What are the general steps for converting a double integral to polar coordinates? 2. How do you determine the limits of integration when switching to polar coordinates? 3. What is the Jacobian determinant and why does $$dx\,dy$$ become $$r\,dr\,d\theta$$ in polar coordinates? 4. How would the solution change if the region of integration were a semicircle instead of a quarter-circle? 5. Could the same problem be solved directly in Cartesian coordinates? If so, how? ### One Tip: When converting to polar coordinates, always sketch the region of integration to correctly identify the limits for $$r$$ and $$\theta$$.

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Double Integration
Polar Coordinates
Jacobian Transformation

x = r cos(θ), y = r sin(θ)
x^2 + y^2 = r^2
dx dy = r dr dθ

Transformation of coordinates
Evaluation of integrals over circular regions

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