Question

Sketch the region \(R\) and evaluate the iterated integral \(\int_{R} \int f(x, y) d A .\) $$ \int_{-a}^{a} \int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2}-x^{2}}}(x+y) d y d x $$

Short answer

The region is a circle centered at the origin with radius \( a \), and the value of the iterated integral over this region is \( 4a^{3}/3 \).

Step-by-step solution

Sketch Region

The region corresponds to the value of \(\ R \) which lies between \( x = -a \) and \( x = a \), and \( y = -\sqrt{a^{2}-x^{2}} \) and \( y = \sqrt{a^{2}-x^{2}} \). This represents a circle centered at the origin with radius \( a \) in the xy-plane.

Change Order of Integration

The integral will be easier to evaluate if we change the order of integration to dy dx. When we do this, the limits become \( y = -a \) to \( y = a \) and \( x = -\sqrt{a^{2}-y^{2}} \) to \( x = \sqrt{a^{2}-y^{2}} \).

Evaluate Inner Integral

Integrate the inner integral first, \( \int_{-\sqrt{a^{2}-y^{2}}}^{\sqrt{a^{2}-y^{2}}} (x + y) \, dx \). This results in \( [x^{2}/2 + yx]_{-\sqrt{a^{2}-y^{2}}}^{\sqrt{a^{2}-y^{2}}} \), which simplifies to \( a^{2}-y^{2} + 2ay \).

Evaluate Outer Integral

Now we can integrate \( \int_{-a}^{a} (a^{2}-y^{2} + 2ay) \, dy \). The result is \( [a^{2}y - y^{3}/3 + ay^{2}]_{-a}^{a} \), which simplifies to \( 4a^{3}/3 \).

Key concepts

Sketching Regions in Calculus

Understanding the geometric representation of regions is crucial when dealing with double integrals in calculus. When you're given a problem like the one presented, which involves sketching the region for the iterated integral \\(\int_{R} \int f(x, y) dA\)\, the first step is visualizing the area over which you'll integrate. The given integral \\(\int_{-a}^{a} \int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2}-x^{2}}}(x+y) dy dx\) defines a region \(R\) on the \(xy\)-plane.

Visualizing this region involves recognizing the limits of integration as descriptions of a boundary. Here, the limits for \(y\) are \(\pm\sqrt{a^{2}-x^{2}}\), which are the top and bottom halves of a circle with radius \(a\). This circle is centered at the origin because the limits of \(x\) span from \(\-a\) to \(a\), covering the full diameter of the circle in the horizontal direction. Sketching this out, you create a circular region on the \(xy\)-plane.

Being able to sketch regions accurately is a foundational skill in calculus, as it not only helps to visualize the problem but also assists in determining if changing the order of integration might simplify the computation, as we will see in the next section.

Changing Order of Integration

Changing the order of integration is a technique used for simplifying the evaluation of double integrals, like in our exercise. An integral in the form \(\int \int f(x, y) dy dx\) suggests we integrate with respect to \(y\) first, and then \(x\). However, sometimes it's advantageous to switch this order to integrate with respect to \(x\) first.

When we switch the order for our circular region from \(dy dx\) to \(dx dy\), we also change the integration limits to reflect this new order. This means we need to describe \(x\) in terms of \(y\), which flips our interpretation of the circle's equation, resulting in the new limits: \(y = -a\) to \(y = a\), and \(x = -\sqrt{a^{2}-y^{2}}\) to \(x = \sqrt{a^{2}-y^{2}}\).

Performing this switch can often reveal an easier path to finding the integral, as it may simplify the function or the limits of integration. That's exactly what happens in the given problem, allowing for a straightforward integration process once the order is adjusted.

Evaluating Double Integrals

The final step in solving our problem involves the actual evaluation of the double integral. Double integrals allow us to calculate the volume under a surface over a given region \(R\). This is done by integrating a function \(f(x, y)\) over the region in a two-step process - hence 'iterated' integrals.

In our example, once the order of integration has been switched, we evaluate the inner integral first by integrating the function with respect to \(x\), while \(y\) is treated as a constant. After finding the antiderivative and applying the limits of \(x\), we then proceed to the outer integral, integrating the resulting expression with respect to \(y\).

The result of the outer integral gives us the total volume under the surface described by the function \(f(x, y) = x + y\) over the circular region \(R\). It's the accumulation of all the small slices of volume over the particular range of \(y\) and \(x\) we have defined. In this exercise, it simplifies to \(\frac{4a^3}{3}\), giving us the net volume under the curve within the specified limits. This process showcases the beauty of iterated integral calculus in determining multidimensional areas and volumes.