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:
$$y=\sqrt{x}, \space 0&lt;x&lt;9$$
$$x=y^2,\space 0&lt;y&lt;3$$
$$A(y)=\pi x^2=\pi y^4$$
$$V(y)=\int_0^3 \pi y^4\,dy$$
$$V=\frac{243 \pi}{5}$$
$$V(\text{adjusted})=\sin(60^o)*V=\frac{243\sqrt{3 \pi}}{10}$$
Some help to get?
Image:
Notice that if we reposition our axis to be parallel and perpendicular to the water level this is the equaivalent of performing the following:
$$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}\cos(60)&amp;-\sin(60)\\\sin(60)&amp;\cos(60)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$
so you can substituted these into the equation to get new formulas for the lines and then work out from there. Alternatively you could imagine the current axis at normal positions and notice the water is at a 60 degree angle and form new limits to the integral for this
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.
$$f_r(x) = \sqrt{x}$$
$$\int_0^{9 - \sqrt{3}} \pi f_r(x)^2 dx$$
$$\int_0^{9 - \sqrt{3}} \pi x \ dx$$
$$ \pi [\frac{x^2}{2} ]^{9 - \sqrt{3}}_0 $$
$$ \pi * \frac{(9-\sqrt{3})^2}{2} $$
$$=~ 82.974...$$