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?
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 $$V=\int_0^{2\pi}\int_0^{7.7}\int_0^{\frac{101}{1448}r\cos\theta+ 55.86}r\mathbb d\theta\mathbb dr\mathbb dz=$$ $$=\int_0^{2\pi}\int_0^{7.7}[z]_0^{\frac{101}{1448}r\cos\theta+ 55.86}r\mathbb d\theta\mathbb dr=\int_0^{2\pi}\int_0^{7.7}(\frac{101}{1448}r\cos\theta+ 55.86)r\mathbb d\theta\mathbb dr=$$ $$=\int_0^{2\pi}[\frac{101}{3·1448}r^3\cos\theta+\frac{55.86}{2}r^2]_0^{7.7}\mathbb d\theta=$$ $$=\int_0^{2\pi}\left(\frac{101}{3·1448}7.7^3\cos\theta+\frac{55.86}{2}7.7^2\right)\mathbb d\theta=$$ $$=\left[\frac{101}{3·1448}7.7^3\cos\theta\right]_0^{2\pi}+\left[\frac{55.86}{2}7.7^2\theta\right]_0^{2\pi}$$ $$V=\pi7.7^2·55.86$$ 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.