Question

find the area of the part of the cylinder x^2+z^2=a^2 inside the cylinder y^2=a(x+a).

Answer 1

firstly, I dont really know exactly how to do it because the cylinder is around a different way, now its going up and down the y axis, usually it goes up and down the z axis

Answer 2

nevertheless , the first equation suggests that a is the radius ( a constant ) , thus the second equation doesn't even represent a cylinder.

Answer 3

Since the surface is a portion of a cylinder, we will use cylindrical coordinates. The original cylinder can be parametrized by\[\gamma(\theta, y) = (a\cos\theta, y, a\sin\theta), \quad \theta \in [0, 2\pi),\ y \in \mathbb{R}.\]The portion of the cylinder inside the parabola can be parametrized by the same function by changing the intervals where the variables live. Lets start by finding the limits for \[\theta.\]Because of the symmetries of the surfaces (everything is symmetric with respect to the Y axis) it's a good idea to set\[y = 0,\]then\[\gamma(\theta, 0) = (a \cos\theta, 0, a\sin\theta).\]When\[\theta = \pi\]we get\[\gamma(\pi, 0) = (-a, 0, 0)\]which is the vertex of the parabola, so \[\pi\]belongs to the new interval for\[\theta.\]Then, for any other \[\theta\]the point will be inside the parabola (since for every x greater than -a, with = 0 and any z the point will be inside the parabola). This way, we find that the interval for\[\theta\]is\[\theta \in [0, 2\pi).\]Now lets find the interval where y lives, as a function of\[\theta.\]We already established that y = 0 is inside this interval and that it is symmetrical with respect to y = 0. For a given\[\theta,\]we must find a point on the parabola of the form\[\gamma(\theta, y) = (a \cos\theta, y, a\sin\theta).\]Since this point must lie on the parabola, \[y^2 = a(x+a) \Rightarrow y^2 = a(a\cos\theta + a) = a^2(\cos\theta + 1) \Rightarrow y = \pm a\sqrt{\cos\theta + 1},\]so we get \[y \in [-a\sqrt{\cos\theta + 1}, a\sqrt{\cos\theta + 1}].\]The area is\[A = \iint_D dA\]where D is the surface and\[dA = \left\|\frac{\partial \gamma}{\partial \theta} \times \frac{\partial \gamma}{\partial y}\right\|dyd\theta=adyd\theta\]so finally\[A = \int_0^{2\pi}\int_{-a\sqrt{\cos\theta + 1}}^{a\sqrt{\cos\theta + 1}}dyd\theta = 2a\int_0^{2\pi}\sqrt{\cos\theta + 1}d\theta = 8a\sqrt{2}.\]