Question

Evaluate the following integrals using the method of your choice. A sketch is helpful. $$\int_{-4}^{4} \int_{0}^{\sqrt{16-y^{2}}}\left(16-x^{2}-y^{2}\right) d x d y$$

Short answer

In summary, the value of the double integral of the function \(f(x, y) = 16 - x^2 - y^2\) over the given region defined by \(x\) going from \(0\) to \(\sqrt{16 - y^2}\) and \(y\) going from \(-4\) to \(4\) is equal to \(128\pi - \frac{512\pi}{45}\).

Step-by-step solution

Identify the region we're integrating over

First, we need to understand the region we're integrating over. The limits of integration are \(\int_{-4}^{4} \int_{0}^{\sqrt{16-y^{2}}}\), which indicates that we are considering a region in the xy-plane that is bounded by the curve \(x^2 + y^2 = 16\) in the first and second quadrants because \(x\) is only allowed to be non-negative. This region is a semi-circle with radius 4 centered at the origin.

Evaluate the integral with respect to x

We'll start by evaluating the inner integral with respect to x. To integrate the function \((16-x^2-y^2)\) with respect to \(x\), we treat \(y\) as a constant. $$\int_{0}^{\sqrt{16-y^{2}}}(16-x^2-y^2)dx$$ The antiderivative of \(f(x,y)\) with respect to x is: $$16x - \frac{1}{3}x^3 - y^2x$$ Now we'll evaluate this antiderivative at the bounds of integration, \(0\) to \(\sqrt{16-y^2}\): $$\left[16\sqrt{16-y^2} - \frac{1}{3}\left(\sqrt{16-y^2}\right)^3 - y^2\sqrt{16-y^2}\right] - [0]$$ Simplifying this expression, we get: $$16\sqrt{16-y^2} - \frac{1}{3}(16-y^2)^{\frac{3}{2}} - y^2\sqrt{16-y^2}$$

Evaluate the integral with respect to y

Now we need to integrate the expression found in step 2 with respect to y over the interval from \(-4\) to \(4\): $$\int_{-4}^{4} \left(16\sqrt{16-y^2} - \frac{1}{3}(16-y^2)^{\frac{3}{2}} - y^2\sqrt{16-y^2}\right)dy$$ To evaluate this integral, it's best to use polar coordinates. We will transform the expression above into polar coordinates, and then integrate it with respect to \(r\) and \(\theta\). Expressing the integral in polar coordinates, we have: $$\int_{0}^{2\pi} \int_{0}^{4}\left(16r - \frac{1}{3}(16-r^2)^{\frac{3}{2}} - r^3\cos^2\theta\right)r dr d\theta$$ Note that we are integrating over the region defined by \(\theta\) going from \(0\) to \(2\pi\) and r going from \(0\) to \(4\). This is because we are integrating over a full circle with radius 4.

Evaluate the integral with respect to r and θ

We'll first integrate with respect to \(r\) and then with respect to \(\theta\). Integrating with respect to \(r\) yields: $$\int_{0}^{2\pi}\left(\left[8r^2 - \frac{1}{9}\cdot\frac{2}{5}\cdot(16-r^2)^{\frac{5}{2}} - \frac{1}{4}r^4\cos^2\theta\right]_0^4 \right)d\theta$$ Evaluating this expression at the bounds r = 0 and r = 4 gives: $$\int_{0}^{2\pi}\left(128-\frac{256}{45}-64\cos^2\theta\right)d\theta$$ Now, we will integrate with respect to \(\theta\): $$\int_{0}^{2\pi}\left(128-\frac{256}{45}-64\cos^2\theta\right)d\theta$$ Integrating with respect to \(\theta\) and evaluating at the bounds yields: $$\left[\left(128\theta-\frac{256}{45}\theta-32\sin{2\theta}\right)_0^{2\pi}\right]$$ Evaluating this expression gives us the final result: $$\boxed{128\pi-\frac{512\pi}{45}}.$$

Key concepts

Polar Coordinates

Polar coordinates provide a useful way to describe points in the plane using a radius and an angle. In polar coordinates, any point is defined by its distance from a fixed point (usually the origin) and by the angle between the line connecting the point to the origin and a fixed direction (usually the positive x-axis). This is particularly useful when dealing with circular regions, as it simplifies the integration process.

For example, in the solution, the expression in Cartesian coordinates is transformed to polar. This transforms region constraints like a semi-circle, bounded by equations of the form \(x^2 + y^2 = R^2\), into a simpler form where \(r\) is merely the radius and \(\theta\) the angle, which simplifies to \(0 \leq r \leq R\) and \(0 \leq \theta \leq 2\pi\).

• It's effective in integrating over spherical regions.
• Simplifies integration bounds into more manageable limits.
• Requires understanding of trigonometric conversions that relate Cartesian and polar coordinates, such as \(x = r\cos\theta\) and \(y = r\sin\theta\).

Integration Bounds

Integration bounds are the limits that define the region over which we integrate. In polar coordinates, these are often specified as ranges for \(r\) and \(\theta\), as opposed to Cartesian coordinates which use \(x\) and \(y\). Proper understanding of the limits is crucial to correctly evaluating an integral.

In this problem, the integration bounds were first interpreted in Cartesian coordinates, forming a semi-circle which simplified to a circular region after conversion to polar coordinates.

• In Cartesian form, limits are more complex (\(x\) from 0 to \(\sqrt{16-y^2}\) and \(y\) from -4 to 4).
• When converted to polar form, limits become straightforward (\(r\) from 0 to 4, \(\theta\) from 0 to \(2\pi\)).
• Identifying and understanding these bounds accurately is essential for the integrity of the solution.

Antiderivatives

Antiderivatives play a vital role in the evaluation of definite integrals. An antiderivative, in simple terms, is the reverse of a derivative. It's used to find the original function whose derivative yields the integrand.

In this exercise, after setting up the bounds and region, we find the antiderivative of the expression \(16 - x^2 - y^2\) with respect to \(x\). This results in \(16x - \frac{1}{3}x^3 - y^2x\). This expression is then evaluated at the integration limits to continue solving the integral.

• Solving involves taking the antiderivative of the function with respect to one variable first.
• The process is repeated for the second variable after substituting back the evaluated antiderivative.
• It's necessary to properly apply constant rules and power rules for integration.

Region of Integration

The region of integration refers to the area or volume in space over which we integrate the function. In double integrals, this region potentially includes two variables, such as \(x\) and \(y\). Understanding the area being evaluated is foundational to setting up correct bounds for the integration process.

Initially, the region of integration in this problem was a semi-circle described by \(x^2 + y^2 = 16\), limited to part of the Cartesian plane. However, when this region is converted to polar coordinates, it becomes a full circle.

• A sketch can be helpful to visually comprehend the change of region after changing coordinates.
• Converting to polar coordinates simplifies the process when dealing with circular or radial symmetry.
• Knowing how to switch between Cartesian and polar interpretations will strengthen your problem-solving skills in integration.