\(\displaystyle{ -4^3 \left[ \frac{\sqrt{16-x^2}}{4} - \right. }\) \(\displaystyle{ \left. \frac{1}{3} \left( \frac{\sqrt{16-x^2}}{4} \right)^3 \right] + C }\)

\(\displaystyle{ -16\sqrt{16-x^2} \left[ 1-\frac{1}{48}(16-x^2) \right] + C }\)

\(\displaystyle{ -(1/3)\sqrt{16-x^2} (x^2+32) + C}\)

\(\displaystyle{ \frac{x+2}{\sqrt{9-(x+2)^2}} + \frac{1}{3}\left[ \frac{x+2}{\sqrt{9-(x+2)^2}} \right]^3 + C }\)

\(\displaystyle{ \frac{x+2}{3\sqrt{9-(x+2)^2}} }\) \(\displaystyle{ \left[ 3 + \frac{(x+2)^2}{9-(x+2)^2} \right] + C }\)

\(\displaystyle{ \frac{x+2}{3[9-(x+2)^2]^{3/2}} \left[ 3(9-(x+2)^2) + (x+2)^2 \right] + C }\)

\(\displaystyle{ \frac{(x+2)[27-2(x+2)^2]}{3[9-(x+2)^2]^{3/2}} + C }\)

It is debatable whether or not multiplying out the numerator would simplify the answer more. We will leave it factored. However, check with your instructor to see what they would expect.

Our first inclination might be to separate this integral into two integrals by distributing the square root term across \((2-y)\). Although this will work, it takes longer and you end up with 3 integrals to evaluate, not just 2. So, let's simplify the term under the square root first using integration by substitution with \(u=y-3\).

\(u=y-3 \to du = dy\)

\(u=y-3 \to u+3 = y\)

When \(y=0\), \(u=y-3 \to u = -3\)

When \(y=2\), \(u=y-3 \to u=-1\)

Now we will rewrite the integral in terms of \(u\).

\(\displaystyle{ \int_{-3}^{-1}{ (-u-1)\sqrt{9-u^2} ~ du} }\)

To reduce the number of negative signs, we can simplify this integral.

\(\displaystyle{ \int_{-1}^{-3}{ (u+1)\sqrt{9-u^2} ~ du} }\)

That last step is not necessary but losing negative signs is one of the most common errors when evaluating integrals. So that step removed \(-1\) from \((-u-1)\) and absorbed into the integral by switching the limits of integration.

Now is the time to distribute the square root term across \((u+1)\) and split the integral into two integrals.

\(\displaystyle{ \int_{-1}^{-3}{ u \sqrt{9-u^2}~du } + }\) \(\displaystyle{ \int_{-1}^{-3}{ \sqrt{9-u^2} ~du } }\)

Each integral requires different techniques. So let's evaluate them separately. The first one is the easiest, so let's start with that one.

\(\displaystyle{ \int_{-1}^{-3}{ u \sqrt{9-u^2}~du } }\)

Use basic substitution with \(x=9-u^2\). We usually use the variable \(u\) for substitution but we already have a \(u\) in the integral. So we chose to use \(x\) since it had not been used in this problem yet. Otherwise we might have chosen \(v\).

\(x=9-u^2 \to dx = -2u~du\)

when \(u=-1\), \(x=9-u^2 \to x = 8\)

when \(u=-3\), \(x=9-u^2 \to x= 0\)

Using these substitutions, the first integral is

\(\displaystyle{ \int_8^0{ x^{1/2} \frac{dx}{-2} } = }\) \(\displaystyle{ (1/2) \int_0^8{ x^{1/2} ~dx } = }\) \(\displaystyle{ \frac{1}{2} \left[ \frac{x^{3/2}}{3/2} \right]_0^8 = \frac{16\sqrt{2}}{3}}\)

Okay, so now let's work on the second integral.

\(\displaystyle{ \int_{-1}^{-3}{ \sqrt{9-u^2} ~du } }\)

For this integral we need to use trig substitution with \(u = 3\sin\theta\).

\(u = 3\sin\theta \to du = 3\cos \theta ~ d\theta\)

Rather than changing the limits of integration in terms of \(\theta\), we will just drop them to make sure our notation is correct and then convert back to \(u\) before evaluating the limits of integration. Using the substitution our integral is now

\(\displaystyle{ \int{ \sqrt{9-9\sin^2\theta}(3\cos \theta) ~ d\theta } }\)

\(\displaystyle{ 9 \int{ \cos^2 \theta ~ d\theta } }\)

\(\displaystyle{ 9 \int{ \frac{1+\cos(2\theta)}{2} ~ d\theta } }\)

\(\displaystyle{ (9/2) \int{ 1+\cos(2\theta) ~ d\theta } }\)

\(\displaystyle{ \frac{9}{2}\left[ \theta + \frac{\sin(2\theta)}{2} \right] }\)

Since we need to convert this result back in terms of \(u\), use the identity \(\sin(2\theta) = 2\sin \theta \cos \theta \)

The original substitution was \(u=3\sin\theta\). Using this we determine that \( \theta = \arcsin(u/3)\) and \( \cos\theta = (1/3)\sqrt{9-u^2} \). Now we have

\(\displaystyle{ \frac{9}{2} \left[ \arcsin(u/3) + \frac{u}{3} \cdot \frac{\sqrt{9-u^2}}{3} \right] }\)

Okay we are making good progress. Now we need to take the last equation and evaluate it from \(-1\) to \(-3\). Doing this gives us the expression

\(\displaystyle{ \frac{9}{2}\left[ \arcsin(-1) + (-1)(0) \right] - }\) \(\displaystyle{ \frac{9}{2}\left[ \arcsin(-1/3) + \frac{-1}{3}\frac{\sqrt{8}}{3} \right] }\)

\(\displaystyle{ \frac{-9\pi}{4} + \frac{9}{2}\arcsin(1/3) + \sqrt{2} }\)

Okay, that last expression is the result from the second integral. Adding it to the result from the first integral will give us the final answer.