What is wrong in my method of solving the volume of a solid of uniform cross-section A solid of a uniform cross-section of right-angled isosceles triangles perpendicular to the $x$-axis has a base $x^2 + y^2 = 16$. Solve for the volume.
Let $\theta$ be the angle between the $x$-axis and the radii of a circle for $0 \leq \theta \leq \frac{\pi}{2}$. The hypotenuse of the right-angled triangles is $8\sin \theta$. As the relation between a hypotenuse and area of a right-angled isosceles triangle is $\frac{\text{Hyp}^2}{4}$, the area of a single triangle is $16\sin^2 \theta$. As the solid is symmetric, we integrate over the domain and multiply the answer by 2.
\begin{align*}
&\quad\quad 2\int_0^\frac{\pi}{2}16\sin^2 \theta \quad d\theta\\
&= 32\int_0^\frac{\pi}{2}\frac{1}{2}-\frac{1}{2}\cos 2\theta \quad d\theta\\
&=32\left[\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta \right]_0^\frac{\pi}{2}\\
&= 8\pi
\end{align*}
However, this answer is incorrect as when one uses the more conventional method to solve:
Let the hypotenuse of one of the triangles be $2y$. Then the area is $y^2$. 
\begin{align*}
Volume &= \int_{-4}^4y^2dx\\
&= \int_{-4}^416-x^2dx\\
&= \left[16x-\frac{1}{3}x^3\right]_{-4}^4\\
&= \frac{256}{3}
\end{align*}
which appears to be the correct answer. I'm curious as to why my alternative method doesn't work. 

Using a cartesian coordinate system, we have (as you pointed out in your question): 
\begin{equation} 
V = \int_{a}^{b} A(x) dx = \int_{-4}^{4} \frac{(\sqrt2y)(\sqrt2y)}{2} dx = \int_{-4}^{4} y^2 dx = \int_{-4}^{4} 16-x^2 dx = \frac{256}{3}
\end{equation}
Now if you want to use polar coordinates (where $ \theta $ is the angle between the $ x $ axis and the radius of the circle), you should proceed as follows:
\begin{equation} 
V = \int_{a}^{b} A(x) dx = \int_{-4}^{4} y^2 dx 
\end{equation}
Since $ y = 4sin(\theta) $ and $ x = 4cos(\theta) \rightarrow dx = -4sin(\theta)d\theta $, we get:
\begin{equation} 
\int_{cos^{-1}\big(\frac{-4}{4}\big)}^{cos^{-1}\big(\frac{4}{4}\big)} (4sin(\theta))^2 (-4sin(\theta))d\theta = -4^3 \int_{\pi}^{0} sin^3(\theta) d\theta = 4^3 \int_{0}^{\pi} sin^3(\theta) d\theta = \frac{256}{3}
\end{equation}
