I am currently reading Tom Apostol's Calculus Vol.1 and I am somehow stuck on this problem: Having defined $$\pi = 2 \int_{-1}^{1}{\sqrt{1-x^2}dx}~~,$$ use the properties of the integral to compute the following in terms of $~\pi~$: $$\int_{-3}^{3}{\sqrt{9-x^2}dx}$$ The properties that have been shown so far in this book are: $$\int_a^b{\sum_{k=1}^n{c_kf_k(x)dx} = \sum_{k=1}^nc_k\int_a^bf_k(x)dx}$$ $$\int_a^b{f(x)dx}+\int_b^c{f(x)dx} = \int_a^c{f(x)dx}$$ $$\int_a^b{f(x)dx}=\int_{a+c}^{b+c}{f(x-c)dx}$$ $$\int_a^b{g(x)dx} \leq \int_a^b f(x)dx\;\text{If}\; g(x) \leq f(x)$$ Along with proofs of these theorems.
Edit from OP: I have forgotten a property like a dummy. The missing one is: $$\int_a^b{f(x)}dx={\frac{1}{k}\int_{ka}^{kb} f(\frac{x}{k})dx}$$ Sorry for the mistake.
Now since the integrand here is represented by a semi-circle of radius $~3~$, it is fairly obvious that the anwser is $~\frac{9\pi}{2}~$ .
Yet I can't seem to be able to use the aforementioned properties to find it in the way that author wants me to.
I assumed that I could somehow derive the following formula : $$9\int_{-1}^1\sqrt{1-x^2}dx = \int_{-3}^{3} \sqrt{9-x^2}dx$$ And it would follow that the anwser was $\frac{9\pi}{2}$ ,but I keep hitting dead ends.
Could somebody please explain to me the method (involving only the aforementioned properties) that should be used to solve this?
I've been sitting on this for a while and I'm wondering if this is simply a case of not being able to find the right numbers or if I am simply not on the right track.
P.S.: This is my first post here/first time using Mathjax, so please excuse/point out any mistakes I might have made in the formulas.