Double Integral Gone Wrong So, I have the (seemingly) algebraically innocuous double integral of $$ \iint \limits_R{1- {x^2 \over 4} -{y^2 \over 9}\space \mathrm {d}A} ; \space\mathrm {where}\space R =[-1,1]\times[-2,2]  $$ 
I integrate with  by splitting up the statement into the integrals of the terms and then proceed to integrate with respect to $y$. A few simple operations later and this yields $ {20 \over 9} -x^2 $; my subsequent integration with respect to $x$ leads to ${40\over9}-{1\over3} {x^3}]^{1}_{-1}$, which in turn leads to an answer of $34\over9$.
I have no idea what I'm doing wrong here, but this has no obvious relationship to the answer of $166\over27$.
I should also note that the original problem was not so explicit about $A$; the equation I am operating on was given as $$z+ {x^2 \over 4} + {y^2 \over 9} = 1$$
I have verified that $R$ lies completely under the region of integration by computing the values of the four corners of $R$ ( $[-1,-2];[-1,2];[1,-2];[1,2]$) and finding the resultants all to be less than one.
This post is getting long, but if anyone wants to see them I'll latex in my steps here with the single $ operate to keep the writing more compact and the \\ operator to facilitate newlines.
 A: $$\begin{align}\int_{-2}^2\left(1-\frac{x^2}4-\frac{y^2}9\right)\,dy &= \left[y-\frac{x^2}4y-\frac{y^3}{27}\right]_{y=-2}^2\\ &= \left[2-\frac{x^2}4\cdot 2-\frac{2^3}{27}\right]-\left[-2-\frac{x^2}4\cdot-2-\frac{-2^3}{27}\right]\\ &= 2\left[2-\frac{x^2}4\cdot 2-\frac{2^3}{27}\right]\\ &= 2\left[\frac{54}{27}-\frac8{27}-\frac{x^2}2\right]\\ &= 2\left[\frac{46}{27}-\frac{x^2}2\right]\\ &= \frac{92}{27}-x^2.\end{align}$$
A: The integration over $y$ should give $\frac{92}{27}-x^2$ and the final result is indeed $\frac{166}{27}$. I can't tell you where, but you must have made some error in your calculation.
A: The first calculation is wrong. The result should be $\frac{92}{27}-x^2$.
By the way, I would take advantage of symmetry and integrate over $0\le x\le 1$, $0\le y\le 2$, and then multiply by $4$.
A: For a rectangular region, just integrate with respect to one variable first, treating the other as constant:
$$\begin{align}\iint_A dA \left ( 1 - \frac{x^2}{4} - \frac{y^2}{9}\right) &= \int_{-1}^1 dx \: \int_{-2}^2 dy \: \left ( 1 - \frac{x^2}{4} - \frac{y^2}{9}\right)\\ &=  \int_{-1}^1 dx \: \left [ y -\frac{x^2}{4} y - \frac{y^3}{27} \right ]_{-2}^2\\ &= \int_{-1}^1 dx \: \left ( 4 - x^2 - \frac{16}{27}\right )\\&=\int_{-1}^1 dx \: \left ( \frac{92}{27} - x^2\right )\end{align}$$
You should be able to take it from here.
