Let S be the solid with flat base, whose base is the region in the xy plane defined by the curves $$y=e^x$$,$$y=−3$$,$$x=0$$ and $$x=1$$, and whose cross-sections perpendicular to the x axis are semi-circles with diameters that sit in the xy plane.
The question is to find the area A(x) of the cross-section of S given by the semi-circle that stands perpendicular to the xy plane, at coordinate x. How will the half-rotation be and about which axis? (It certainly has to be a half-rotation so that we obtain a flat surface).
1 Answer
$\begingroup$There is no axis of rotation - it's not a surface of revolution.
Imagine slicing the surface at $x = x_0$, somewhere between $0$ and $1$. By definition, this yields a semicircle whose diameter lies on the base. Let's determine the length of this diameter.
The diameter is the line segment whose endpoints are $(x_0, e^{x_0})$ and $(x_0, -3)$. So its length is $e^{x_0} + 3$. By computing the radius and applying the usual semicircle area formula, we obtain: $$ A(x) = \frac{1}{2}\pi(\tfrac{e^x + 3}{2})^2 = \frac{\pi}{8}(e^x + 3)^2 $$
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