I have a given polynomial, f(x)=c1*x^7+c2*x^6+c3*x^5+c4*x^4+c5*x^3+c6*x^2+c7*x+c8 rotated about the y axis, which results in a nice surface:
What I would like to be able to do is evaluate the volume of a portion of this solid for a given square area, say a square defined by the coordinates (200,200) (200,400) (400,200) (400, 400) for the coordinates in the image. Basically the portion of a rectangle of arbitrary height as long as the height is greater than the maximum value reached by the curve:
I have a sneaking suspicions I'm wildly overthinking a somewhat straightforward calculus problem but throwing in coordinate systems and the rotated curve is throwing me off.
The rotated function you're trying to integrate is
$$f(r,\theta) = \sum_{k=0}^n c_k r^k$$
For your case, $n=7$, so the fully written out form would be
$$f(r,\theta) = c_0 + c_1r + c_2r^2+c_3r^3+c_4r^4+c_5r^5+c_6r^6+c_7r^7$$
Note that these $c_i$ are not the same as yours, you can compare the above expression to your own to see which coefficients are which.
To convert to rectangular coordinates we set $r = (x^2+y^2)^\frac{1}{2}$, so
$$f_{\text{rect}}(x,y)= \sum_{k=0}^n c_k (x^2+y^2)^\frac{k}{2}$$
Plugging in $n=7$ and evaluating the integral term-by-term you should be able to get an exact solution,
$$ \sum\limits_{k=0}^7 I_k $$
I'll give an example of an even and an odd term because they have different ways of evaluating.
First, the easier one is the even coefficient, since the term
$$(x^2+y^2)^\frac{k}{2} = (x^2+y^2)^i$$
is a just a polynomial when $k$ is even. Take $k=6$ for instance. The term we need to integrate is then
$$(x^2+y^2)^3 = x^6 + 3x^4y^2 + 3x^2y^4 + y^6$$
This expansion can be done by hand, I won't go into the details of shortcuts but you can search for "binomial expansion shortcut" for some nice tricks. Otherwise, you can write out
$$(x^2+y^2)(x^2+y^2)(x^2+y^2)$$
and use the distributive property.
To integrate, we make use of the fact that integrals are linear (so we can distribute among sums and factor out constants) and a nice property of double integrals
$$\iint f(x)g(y)\text dx \text dy = \int f(x) \text dx \int g(y) \text dy$$
which lets us integrate any polynomial in any number of variables. Very nice.
Here's the full calculation for the $k=6$ term:
$$ \begin{equation}\begin{split} I_6 & = \iint c_6(x^2+y^2)^\frac{6}{2}\text dx \text dy \\ & = \iint c_6(x^2+y^2)^3\text dx \text dy \\ & = c_6 \iint (x^2+y^2)^3\text dx \text dy \\ & = c_6 \iint x^6 + 3x^4y^2 + 3x^2y^4 + y^6\text dx \text dy \\ & = c_6 \left(\iint x^6\text dx \text dy + 3 \iint x^4y^2\text dx \text dy + 3 \iint x^2y^4\text dx \text dy + \iint y^6\text dx \text dy\right) \\ & = c_6 \left(\int x^6\text dx \int \text dy + 3 \int x^4 \text dx \int y^2\text dy + 3 \int x^2 \text dx \int y^4 \text dy + \int \text dx \int y^6 \text dy\right) \\ & = c_6 \left(\frac{x^7 y}{7} + \frac{x^5 y^3}{5} + \frac{x^3 y^5}{5} + \frac{xy^7}{7}\right) \\ \end{split}\end{equation} $$
You can evaluate the last step with limits instead to get the numerical values you're looking for.
Okay, now the odd coefficients. We'll use 1 for the example. The basic idea is to take an integral that looks like
$$\iint (x^2+y^2)^\frac{1}{2} \text dx \text dy$$
and make a substitution $x = y\tan \theta$, so $\text dx = y\sec^2 \theta \text d\theta$. Since $\tan^2 \theta + 1 = \sec^2 \theta$, this integral becomes
$$\iint y^{2}\sec^{3} \theta\text d\theta \text dy$$
using the neat property we used before, this is just
$$\int y^{2}\text dy \int \sec^{3} \theta \text d\theta$$
The first part evaluates easily to $\frac{y^3}{3}$, the second part takes some work. We'll make use of integration by parts with $u = \sec \theta$ and $v = \sec^2 \theta \text d\theta$. This gives $\text du = \sec \theta \tan \theta \text d\theta$ and $v = \tan \theta$. Plugging into the integration by parts formula and simplifying we get
$$\int \sec^{3} \theta \text d\theta = \sec \theta \tan \theta - \int \sec^{3} \theta \text d\theta + \int \sec \theta \text d\theta$$
The integral we're looking for appears on both sides so we can solve for it, and using the known integral $\int \sec \theta \text d\theta$ (can be evaluated by substitution), we have
$$\int \sec^{3} \theta \text d\theta = \frac{1}{2}(\sec \theta \tan \theta + \log|\sec \theta + \tan \theta|)$$
and so finally
$$ I_1 = \frac{1}{6}y^3(\sec \theta \tan \theta + \log|\sec \theta + \tan \theta|) $$
You can reverse the substitution then plug in limits, or you can convert the $(x,y)$ limits to $\theta$ and evaluate that way.
Note that for higher odd powers, you will have to repeat the process of integration by parts and solving for your integral multiple times to reduce to the known integral of $\sec \theta$. You can find some formulae that may help by searching for "integral of odd powers of secant"