Find the volume $V$ of the described solid $S$. The base of $S$ is an elliptical region with boundary curve $9x^2 + 25y^2 = 225$. Cross-sections perpendicular to the $x$-axis are isosceles right triangles with hypotenuse in the base.
I tried this: $$\displaystyle \frac{x^2}{25}+\frac{y^2}{9}=1$$ $$A=\frac{1}{2}l^2(2y^2)=y^2$$ Solving for $y$ I got $$y=\pm3\sqrt{4-x^2}$$ $$V=2\int_{0}^{2}3\sqrt{4-x^2}dx=6\pi$$ But this is wrong. Can you help me?
We find $$y=\pm3\sqrt{1-\frac{x^2}{25}}$$ For a given $x=a$, $y$ represents half the hypotenuse of the triangle built upon $x=a$, and the area of that triangle is $y^2$. Hence the volume of $S$ is $$V=\int_{-5}^5y^2dx$$ $$=\int_{-5}^5\left(3\sqrt{1-\frac{x^2}{25}}\right)^2dx$$ $$=\int_{-5}^59\left(1-\frac{x^2}{25}\right)dx$$ $$=\frac9{25}\int_{-5}^5(25-x^2)dx$$ $$=\frac9{25}\left[25x-\frac{x^3}3\right]_{-5}^5$$ $$=\frac9{25}\left[\frac{250}3-\left(-\frac{250}3\right)\right]$$ $$=60$$