<s>[INST] what is the volume of a tilted glass I was wondering about the volume of a tilted glass (60$^o$).
Not sure, but my attempt:
$$y=\sqrt{x}, \space 0<x<9$$
$$x=y^2,\space 0<y<3$$
$$A(y)=\pi x^2=\pi y^4$$
$$V(y)=\int_0^3 \pi y^4\,dy$$
$$V=\frac{243 \pi}{5}$$
$$V(\text{adjusted})=\sin(60^o)*V=\frac{243\sqrt{3 \pi}}{10}$$
Some help to get?
Image:

 [/INST] The line of the $60$ degree angle can be described by $x = \frac{1}{\sqrt{3}}y + 9 - \sqrt{3}$
The area enclosed by the curve and the line can be calculated with an integral.
$\int{}{}{\frac{1}{\sqrt{3}}y + 9 - \sqrt{3}} - y^2$
You need to find the limits of integration. For the right side, it will be 3 since the limit of the glass is 9. For the left side, it will be $-\sqrt{9-2\sqrt{3}}$.
$$\int_{-\sqrt{9-2\sqrt{3}}}^3{\frac{1}{\sqrt{3}}y + 9 - \sqrt{3}} - y^2 dy$$
Find the value of the integral and that is the two dimensional area of the glass tilted.
To find the volume for a three dimensional version you will have to evaluate an integral of the area of a circle from zero to the liquid level.
$$f_r(x) = \sqrt{x}$$
$$\int_0^{9 - \sqrt{3}} \pi f_r(x)^2 dx$$
$$\int_0^{9 - \sqrt{3}} \pi x \ dx$$
$$ \pi [\frac{x^2}{2} ]^{9 - \sqrt{3}}_0 $$
$$ \pi * \frac{(9-\sqrt{3})^2}{2} $$
$$=~ 82.974...$$
</s>