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I need to calculate the volume between $y = x$ and $x = 4y - y^2$ around the $y$-axis

I am not sure how I can not convert the $x = 4y - y^2$ to being of the form $y = \cdots$ so that I could use it in the shell method: $V = \int 2\pi\times x(x-\cdots) \, dx$ going from $0$ to $3$

Help would be highly appreciated!

Bobson Dugnutt
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Nicolas
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  • Some questions to think about: Have you made a sketch of the region being revolved? Why might you be "forced" to use the shell method, or to solve for $y$ as a function of $x$? What is the geometric difference between shells and washers, and does the difference depend on the names of the axes? – Andrew D. Hwang Mar 11 '16 at 23:06
  • Are you required to use the shell method? You are going to have to break your volume into two integrals: one between the line and the upper "arm" of the horizontal parabola for $ \ 0 \ \le \ x \ \le \ 3 \ $ , the other between the upper and lower "arms" of the parabola for $ \ 3 \ \le \ x \ \le \ 4 \ $ . The disk method, integrating in the $ \ y- $ direction would be less trouble. – colormegone Mar 11 '16 at 23:06
  • The "arms" of the parabola are found by solving the quadratic equation $$ \ y^2 \ - \ 4 \ y \ + \ x \ = \ 0 \ \ \Rightarrow \ \ y \ = \ 2 \ \pm \ \sqrt{4 - x} \ $$ . And in the previous comment, for $ \ 0 \ \le \ x \ \le \ 3 \ $ , I should have written "lower arm" of the parabola. – colormegone Mar 11 '16 at 23:14
  • @RecklessReckoner I solved it using the disk method, that did the trick, thanks! – Nicolas Mar 11 '16 at 23:21

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