Definite integral: $\int^{4}_0 (16-x^2)^{\frac{3}{2}}\,dx$

$\int(16-x^2)^{3/2}=\int(16-x^2)\sqrt{16-x^2}=16\int\sqrt{16-x^2}-\int x^2\sqrt{16-x^2}$. The first integral you can handle. The second one goes by parts: let $u=x$, $dv=x\sqrt{16-x^2}dx$, then antidifferentiate $dv$ by a simple substitution, and you get back to $\int(16-x^2)^{3/2}$.

Looks like you're back where you started, but in fact you get an equation for $\int(16-x^2)^{3/2}$.

A second way to do pretty much the same thing: let $$I=\int(16-x^2)^{3/2}dx$$ Do integration by parts with $u=(16-x^2)^{3/2}$, $dv=dx$, $du=-3x(16-x^2)^{1/2}dx$, $v=x$; $$I=x(16-x^2)^{3/2}+3\int x^2(16-x^2)^{1/2}dx$$ Now $$\int x^2(16-x^2)^{1/2}dx=\int(16-(16-x^2))(16-x^2)^{1/2}dx=16\int(16-x^2)^{1/2}dx-I$$ so we have $$I=x(16-x^2)^{3/2}+48\int(16-x^2)^{1/2}dx-3I$$ from which you get an expression for $I$ in terms of the integral you wanted to use.

If you absolutely only want to use the area of the circle quadrant, you can use Differentiation Under the Integral Sign.

If $$F(a) = \int_{l(a)}^{u(a)} f(a,x) \ \text{d}x$$

then (under appropriate hypotheses) we have that

$$F'(a) = f(a,u(a))u'(a) - f(a,l(a))l'(a) + \int_{l(a)}^{u(a)} \frac{\partial f(a,x)}{\partial a} \ \text{d} x$$

Thus if $$F(a) = \int_{0}^{a} (a^2 - x^2)^{\frac{3}{2}} \ \text{d}x$$

Then since $\displaystyle f(a,a) = (a^2 - a^2)^{\frac{3}{2}} = 0$ we get by applying the above that

$$F'(a) = \int_{0}^{a} \frac{\partial (a^2 - x^2)^{\frac{3}{2}}}{\partial a} \ \text{d}x = 3a \int_{0}^{a} (a^2 - x^2)^{\frac{1}{2}} \ \text{d}x = \frac{3 \pi a^3}{4}$$

(this is where we used that $\displaystyle \int_{0}^{a} (a^2 - x^2)^{\frac{1}{2}} \ \text{d}x$ is the area of a quadrant of the circle of radius $a$).

Since $\displaystyle F(0) = 0$, we have that

$$F(a) = \frac{3 \pi a^4}{16}$$

Thus your integral is $$F(4) = 48 \pi$$

What follows does not really help to evaluate the integral (but please see the comment towards the end). However, there is geometric content.

We look for some geometric insight about $\int_0^4 (16-w^2)^{3/2}dw$, or more generally about $$\int_0^R (R^2-u^2)^{3/2}dw$$ where $R$ is positive.

First we do a dimensional analysis. Suppose that $R$ is measured in feet. Then $(R^2-w^2)^{3/2}$ is measured in cubic feet, as is $dw$, so the integral is measured in feet$^4$. Interesting, looks like the volume of a $4$-dimensional object, maybe a ball. So let's look at the ball of radius $R$ in $4$-dimensional space.

Because of my limited drawing skills, it is convenient to resort to some algebra. The standard ball of radius $R$ has equation $$w^2+x^2+y^2+z^2=R^2.$$ Lets find its volume, maybe using slicing, or cylindrical shells.

Take a slice of thickness $dw$, perpendicular to the $w$-axis, at "height" $w$. The cross-section looks like a sphere. From $$x^2+y^2+z^2=R^2-w^2$$ we can see that the radius of the sphere of cross-section is $(R^2-w^2)^{1/2}$. Thus the hypervolume, if that's a word, of the slice is roughly $$\frac{4\pi}{3}(R^2-w^2)^{3/2} dw.$$ Now add up, from $-R$ to $R$, or just double the integral from $0$ to $R$. We find that the volume of the ball is $V$, where $$V= \int_0^R \frac{8\pi}{3}(R^2-w^2)^{3/2} dw.$$

Now look up the volume of a ball in some standard source. We find that $V=\frac{\pi^2}{2}R^4$ and conclude that $$\int_0^R \frac{8\pi}{3}(R^2-w^2)^{3/2} dw=\frac{3\pi}{16}R^4.$$

Comment: All of the above is useless for evaluating the integral, since the most reasonable way to compute the volume is to evaluate the integral. But there are alternatives.

One of them is described here. Unfortunately it involves ideas more sophisticated than the usual trigonometric substitution or parts. More interestingly, an "elementary" integration-free approach is roughly sketched here.