Find the volume $V$ of the solid bounded by the planes $x+y-z=3$ and $z=0$, and the cylinder $x^2+\frac{y^2}{4}=1$. My calculations give
Polar $$V = \int_{\theta=0}^{\theta=\pi/2} \int_{r=0}^{r=1} \int_{z=r(\cos\theta+\sin\theta)-3}^{z=0} 2r \, \mathrm{d}z \, \mathrm{d}r \, \mathrm{d} \, \theta$$
Rectangular $$V = \int_{x=1}^{x=3} \int_{y=3-x}^{y=\sqrt{4(1-x^2)}} \int_{z=x+y-3}^{z=0} \, \mathrm{d}z \, \mathrm{d}y \, \mathrm{d}x $$
Are these the correct limits?
The cross-section of the cylinder is an ellipse with a semi-axis of $ \ 1 \ $ along the $ \ x-$ axis and a semi-axis of $ \ 2 \ $ along the $ \ y-$ axis. Unfortunately, the inclined plane $ \ z = x + y - 3 \ $ has its symmetry along a line making a 45º angle to the coordinate axes, so we don't catch any break from symmetry in making this volume integration. The "vertical distance" from the tilted plane "upward" to the $ \ xy-$ plane ( $ z = 0 $ ) must be integrated over the full ellipse in the $ \ xy-$ plane, giving us
$$ \ \int_{-1}^{1} \ \int_{-2 \sqrt{1-x^2}}^{+2 \sqrt{1-x^2}} \ \int^0_{x+y-3} \ \ dz \ dy \ dx \ \ . $$
I agree with some of your integration limits, but not all of them. The integration is not as bad as it may initially appear, as one pair of terms cancels out and another term is an odd function of $ \ x \ $ ; the remaining portion of the integration can even be done "geometrically".
I find a value of $ \ 6 \pi \ $ , which Alan confirms in a comment below. As a check, we can determine the minimum and maximum values of $ \ z \ $ over the ellipse, which occur on the curve itself at $ \ ( \ \pm \frac{1}{2 \sqrt{5}} \ , \ \pm \frac{2}{\sqrt{5}} \ ) \ $ . So the extremal values of $ \ z \ $ are $ \ ( \pm \frac{\sqrt{5}}{2}) - 3 \ $ , giving an "average" over the ellipse of $ \ -3 \ $ . Since the area of the ellipse is $ \ \pi \ \cdot \ 1 \ \cdot \ 2 \ $ , we can estimate the volume as $ \ |-3| \ \cdot \ 2 \pi \ = \ 6 \pi \ $ .