Evaluate the volume using only double integrals - possible? I want to find the volume under the slanted plane
$$f(x,y)=\frac{101}{1448}x + 55.86 $$
bounded by a circle with radius 7.7. 
how come 
$$2\int ^\pi _0 \int ^{7.7} _0 [\frac{101}{1448} (r cos \theta) +55.86] \, r \, dr \, d \theta $$
doesn't find the correct volume? Is it because I can only work with symmetrical functions with polar coordinates?
Can someone answer the question with a formula in purely double integrals that can find the volume under that slanted plane? 
 A: Polar coordinates work as any other coordinates. Simply, if the problem has rotation symmetry you can use them. Anyway, the problem is about volume, so you need cylindrical coordinates $(r,\theta,z)$ or cartesian $(x,y,z)$ or whatever with three numbers. But the problem, in part, presents axial symmetry around the z axis, so, cylindrical are presumably easier. Further, if you want the volume, you have to integrate the volume element $r\mathbb d\theta\mathbb dr\mathbb dz$ with the suitable integration limits. Furthermore, you need a triple integral. The limits of integration are:
$r<7.7$; $0<z<\frac{101}{1448}r\cos\theta + 55.86$; $0\le z<2\pi$
$$V=\int_0^{2\pi}\int_0^{7.7}\int_0^{\frac{101}{1448}r\cos\theta+ 55.86}r\mathbb d\theta\mathbb dr\mathbb dz=$$
$$=\int_0^{2\pi}\int_0^{7.7}[z]_0^{\frac{101}{1448}r\cos\theta+ 55.86}r\mathbb d\theta\mathbb dr=\int_0^{2\pi}\int_0^{7.7}(\frac{101}{1448}r\cos\theta+ 55.86)r\mathbb d\theta\mathbb dr=$$
$$=\int_0^{2\pi}[\frac{101}{3·1448}r^3\cos\theta+\frac{55.86}{2}r^2]_0^{7.7}\mathbb d\theta=$$
$$=\int_0^{2\pi}\left(\frac{101}{3·1448}7.7^3\cos\theta+\frac{55.86}{2}7.7^2\right)\mathbb d\theta=$$
$$=\left[\frac{101}{3·1448}7.7^3\cos\theta\right]_0^{2\pi}+\left[\frac{55.86}{2}7.7^2\theta\right]_0^{2\pi}$$
$$V=\pi7.7^2·55.86$$
Wich corresponds exactly to a cylinder of base a circle or radius 7.7 and heigh 55.86, as if no slant exists, as I expected
You expect a different volume because the slant of the top, but there isn't: the center of the top surface (a ellipse, btw) is at $z=55.86$ and, so say, as many volume you add "up the hill" as you substract "down the hill" (you can check it comparing the differences by drawing a "not slanted" cylinder of the same heigh over the "slanted" one). And you expect too a double integral, not triple. I have only a triple one, because the slant "breaks" the symmetry and forces to a triple integral! Anyway, you have a "no integral at all" solution if you want to do what I just suggested: draw a "not slanted" cylinder in the same place as the "slanted" one and compare.
