# Finding the volume of a cone with and oblique base.

The base of $S$ is an elliptical region with boundary curve $9x^2+4y^2=36$. Cross-sections perpendicular to the $x$-axis are isosceles right triangles with hypotenuse in the base. The base of $S$ is an elliptical region with boundary curve $9x^2+4y^2=36$. Cross-sections perpendicular to the $x$-axis are isosceles right triangles with hypotenuse in the base.

$9x^2+4y^2=36 \implies \dfrac{x^2}{2^2}+\dfrac{y^2}{3^2}=1$. Based on the definition of the cross-section, the area of the triangular cross section ($T$) is $T=y^2=\frac{3}{2}\sqrt{4-x^2}$. By the symmetry of the object, the volume of $S$ is $3\int_{0}^{2}\sqrt{4-x^2}dx = V (\text{units}^3)$.

Is this true?

• The solid you describe is not a cone, but it looks good except you have $y^2=\frac{3}{2}\sqrt{4-x^2}$, when actually, $y^2=\frac{9}{4}(4-x^2).$ – Alex S Jun 25 '15 at 4:48
• @AlexS What would the shape be called? Also, I wonder, is there a better method for determining the volume of this object? This way feels rather contrived and extensive. Perhaps just integrating with respect to $y$? Like: $\int_{0}^{3}y^2dy = V$? – alxmke Jun 25 '15 at 4:53
• Nice name. I'm not aware of a name for this shape. As for your other concern, I'm afraid this is the best way (that I've ever seen) to solve this kind of problem. The issue with the integral you suggested is that the area of the triangular region at $y$ is $$\frac{4}{9}(9-y^2),$$ not $y^2$, so the integrand is essentially the same, and integrates to the same value. – Alex S Jun 25 '15 at 5:08
• @AlexS Same to you. You should post your comments as the answer so I can properly credit you! Thanks for the assistance. I'm having quite a time of it with my summer Calculus course. The work is immense. – alxmke Jun 25 '15 at 5:17

You are right that the area of each triangle is $y^2$, but that area is $y^2=\frac{9}{4}(4-x^2).$ Then $$V=\frac{9}{4}\int_{-2}^2(4-x^2)\,dx.$$ (Notice the $-2$ bound on the integral, the ellipse goes from $x=-2$ to $x=2$.)