Hello Airshow4444,
I would first rewrite the given elliptical base in standard form so we may determine the length of the horizontal axis:
$$25x^2+36y^2=900$$
Divide through by $900$:
$$\frac{x^2}{6^2}+\frac{y^2}{5^2}=1$$
Hence, we see the lower limit of integration is $a=-6$ and $b=6$.
The base $B$ of the triangular cross-section for a given $x$ is:
$$B(x)=2y(x)=2\frac{5}{6}\sqrt{6^2-x^2}=\frac{5}{3}\sqrt{6^2-x^2}$$
The height $h$ of the triangular cross-section for a given $x$ is:
$$h(x)=y(x)=\frac{5}{6}\sqrt{6^2-x^2}$$
The area $A$ of the triangular cross-section for a given $x$ is:
$$A(x)=\frac{1}{2}B(x)h(x)=\frac{25}{36}\left(6^2-x^2 \right)$$
Hence, the volume of the solid is given by:
$$V=\frac{25}{36}\int_{-6}^{6}6^2-x^2\,dx=\frac{25}{16}\int_{0}^{6}6^2-x^2\,dx$$
Note: the integrand is an even function, therefore we may apply the even function rule.
Applying the FTOC, we find:
$$V=\frac{25}{18}\left[6^2x-\frac{1}{3}x^3 \right]_0^6=\frac{25}{18}\cdot6^3\left(1-\frac{1}{3} \right)=200$$