Calculus II Professor will not accept my correct integral evaluation that uses a different method, should I bring this up further? I am a freshman enrolled at an American University. Recently, I took an examination in which the following problem appeared:

Evaluate the following integral:

$\int_0^4\sqrt{16-x^2}dx$

My answer: 4$\pi$, was correct. However, I received reduced credit for this answer because I did not solve it correctly (according to the professor). The exams are time-limited and have a fair amount of content, so when I saw this problem, I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $\frac{1}{4}$$\pi$$r^2$ and wrote my answer of 4$\pi$.

The context of the test was surrounding our unit on inverse trigonometry and integration by parts. This section of the test did not list any other instructions besides evaluating the definite integrals. I've talked to my professor about it and his only response was that I solved it wrong:


  To receive full credit, you would have had to evaluate an integral, as the instructions indicated.


Is my interpretation of evaluating the integral different? Does the instruction "Find the antiderivative and then evaluate" not need to exist for that to be required?

Thank you.
I've done some undergraduate teaching and my policy is always if you get the correct answer by any means then you get full credit, but others have different policies and it's really up to them.

You could argue your case.  Your professor could argue back that solving the integral by trig substitution does not require the formula $A = \pi r^2$, and he did not permit the use of that formula.  He could argue that using that formula entails circular reasoning (the formula for the area of a circle has to be gotten by some limiting or integration method equivalent to evaluating $\int_{-r}^r \sqrt{r^2-x^2}dx$).  

It could go either way for you.  But I think it would be a waste of your and your professor's time.
I love your solution! It reminds me of the following question. What is ${{\int}_0}^{2\pi}{\rm{cos}^2}x{\rm{d}}x$? Answer: $\pi$. The reasoning is that as ${\rm{cos}^2}x + {\rm{sin}^2}x = 1$, as the areas under ${\rm{sin}^2}x$ and ${\rm{cos}^2}x$ in the interval $[0,2{\pi}]$ are equal, and as ${{\int}_0}^{2\pi}{{\rm d}}x = 2{\pi}$, we conclude that $${{\int}_0}^{2\pi}{cos^2}x{\rm{d}}x ={{\int}_0}^{2\pi}{sin^2}x{\rm{d}}x = {\pi}$$ The whole point of mathematics to me is beautiful pieces of inspiration like the one you had on your exam. You should get full credit.