What am I doing wrong setting this volume problem up?

Question

The problem states that the base of a solid is the region bounded by the curve $3x^2+y^2=6$. Calculate the volume of the solid if the cross sections perpendicular to the $x$ axis are equilateral triangles with one leg on the base. What I did was: Solve for y and then find the roots of the function which are $\pm\sqrt2$ those are the limits of integration of the area of one triangle $A=2(6-3x^2)$ So the volume is equal to $$V={\displaystyle \int_{-\sqrt2}^{\sqrt2}}2(6-3x^2)dx=2^{\frac{9}{2}}$$ This wrong beacause the answer to the problem is $8\sqrt6$. What am I doing Wrong and how do I solve this problem ?

Answer 1

You calculated the area of the triangle incorrectely. At a given $x$ the range in $y$ is $[-\sqrt{6-3x^2},+\sqrt{6-3x^2}]$. The side of the equilateral triangle is $2\sqrt{6-3x^2}$ and the area of the triangle is $(6-3x^2)\frac {\sqrt 3}2$. The integral should then be $$V=\frac {\sqrt 3}2\int_{-\sqrt 2}^{\sqrt 2}(6-3x^2)dx$$