what is the volume of a tilted glass

Question

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:

Answer 1

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.

$$f_r(x) = \sqrt{x}$$ $$\int_0^{9 - \sqrt{3}} \pi f_r(x)^2 dx$$

$$\int_0^{9 - \sqrt{3}} \pi x \ dx$$ $$ \pi [\frac{x^2}{2} ]^{9 - \sqrt{3}}_0 $$

$$ \pi * \frac{(9-\sqrt{3})^2}{2} $$ $$=~ 82.974...$$

Answer 2

Notice that if we reposition our axis to be parallel and perpendicular to the water level this is the equaivalent of performing the following: $$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}\cos(60)&-\sin(60)\\\sin(60)&\cos(60)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$ 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

Answer 3

We can find the volume without computing integrals, using a theorem by Archimedes of Syracuse:

The volume of a segment of a paraboloid of revolution is $3/2$ the volume of the cone which has the same base and whose vertex lies on the paraboloid, forming with the center of the base a line parallel to its axis.

Note that the plane tangent to the paraboloid at the vertex of the cone is parallel the base. Hence that cone could also be described as the cone of maximal volume inscribed into the paraboloid.

Below you can see a section of the paraboloid, perpendicular to the surface of the water (line $AB$). The base of the segment is an ellipse, having $AB$ as major axis and its midpoint $M$ as center. The cone described in the above theorem has vertex $V$, with $VM$ parallel to the axis of the paraboloid.

It is easy to find the coordinates of those points: $$ A=(9,3);\quad B=\left(\frac{28}{3}-2 \sqrt{3},\frac{1}{\sqrt{3}}-3\right);\quad M=\left(\frac{55}{6}-\sqrt{3},\frac{1}{2 \sqrt{3}}\right);\quad V=\left(\frac{1}{12},\frac{1}{2 \sqrt{3}}\right). $$ The semi-major axis of the base is then $$ a={1\over2}AB=2 \sqrt{3}-\frac{1}{3}, $$ while the semi-major axis is the distance of $M$ from the paraboloid, along a direction perpendicular to plane $ABV$: $$ b=\sqrt{x_M-y_M^2}=3-\frac{1}{2 \sqrt{3}}. $$ Finally, we can compute the height $h$ of the cone: $$ h=VH=VM\sin60°=\frac{109}{8 \sqrt{3}}-\frac{3}{2}. $$ The volume of the segment of paraboloid is thus: $$ {3\over2}\cdot{1\over3}\cdot\pi ab\cdot h = \frac{1}{288} \left(12313-2616 \sqrt{3}\right) \pi \approx 84.887962. $$