Answer to Question #131365 in Calculus for Moel Tariburu

Question

Use cylindrical shells to find the volume of the solid that results when the region enclosed by x = y^2 and x = y is revolved about the line y= -1

Answer 1

We should rotate the region above about the line y=-1.This volume can be represented as the difference between the volume V₁ obtained by rotating the upper curve and the volume V₂ obtained by rotating the lower curve. Two curves intersect when $y^2=y$ in points $(0;0), (1;1)$ .

Every volume can be represented as he sum of volumes of small cylinders with height dx and width f(x)-(-1), so the volume of such an cylinder is $\pi (f(x)+1)^2dx$ Therefore, to obtain V₁ and V₂ we should sum the volumes of cylinders. If we consider smaller and smaller dx, the sum will turn into integral:

$V_1 = \int\limits_0^1 \pi (\sqrt{x}+1)^2dx =\pi \int\limits_0^1 (x+2\sqrt{x}+1)dx= \pi \left(\dfrac{x^2}{2} + \dfrac{4}{3}x^{3/2}+x \right) \Big|^1_0 = \dfrac{17}{6}\pi.$

$V_2 = \int\limits_0^1 \pi (x+1)^2dx = \pi\left(x^3/3+x^2+x\right)\Big|^1_0 = \dfrac{7\pi}{3}.$

The difference between volumes is $\dfrac{\pi}{2}.$