Calculating the Volume of a Curved Hose using Integral Calculus? As similarly shown in the picture, is there way to find the volume of a hose with the same radius throughout the shape using integral calculus?

 A: If you imagine taking infinitesimal slices of the hose, you see that even when the hose curves, the area of a slice stays the same. So the volume is just the arc length times the area of a slice.
Edit: and this arc length is found with the usual arc length integral: $L = \int_a^b \sqrt{\frac{dy}{dt}^2 + \frac{dx}{dt}^2} dt$
A: If we imagine our hose to be parametrized by $\mathbf{x}(s)$ for $s_0 \leq s \leq s_1$ and our volume elements to be discs of radius $R$ centered at $\mathbf{x}(s)$, then our volume integral looks something like $$\int_{s_0}^{s_1} \pi R^2\lvert \dot{\mathbf{x}}(s)\rvert\,\mathrm{d}s$$ Using this we can, for example, compute the volume of a torus with major radius $R$ and minor radius $r$ as follows: Let $\mathbf{x}(\theta) = R\cos(\theta)\,\mathbf{\hat{x}}+R\sin(\theta)\,\mathbf{\hat{y}}$, where $\theta$ ranges between $0$ and $2\pi$. Then, $\lvert \dot{\mathbf{x}}(\theta)\rvert = R$, so the volume of the torus is $$\int_0^{2\pi} \pi r^2R\,\mathrm{\theta} = 2\pi^2r^2R$$ The volume of a cylinder of radius $R$ is similarly easy: Let $\mathbf{x}(s) = hs\,\mathbf{\hat{n}}$, where $h$ is the height of the cylinder, $s$ ranges from $0$ to $1$, and $\mathbf{\hat{n}}$ is the cylinder's central axis. Then, $\lvert \dot{\mathbf{x}}(s)\rvert = h$, so the volume is $$\int_0^1 \pi R^2h\,\mathrm{d}s = \pi R^2h$$
