Find the "volume of the solid that results when the region bounded by $x=1-y^2$ and the y-axis is revolved around the y-axis"

This is from a worksheet that my teacher gave me.

my attempt: $$2\int_0^1 (-x+1) dx$$

What am I doing wrong?


We need to rotate about the line $x=0$ (y-axis), and we want to express $x$ as a function of y: $x=1-y^2$ (as given). That's a horizontal parabola, intersecting the y-axis at $y =-1$ and $y = 1$. The aim here is to integrate with respect to $y$, where $x = 1-y^2$ is the radius, which varies as $y$ varies.

$$\begin{align} \pi \int_{-1}^1 (1-y^2)^2\,dy & = \pi \int_{-1}^1 (1- 2y^2 + y^4)\,dy\\ & = \pi\left(y - \frac {2y^3} 3 + \frac{y^5}5\right)\Bigg|_{-1}^1 \end{align}$$ I'll let you take it from here: just be careful with the signs.

  • $\begingroup$ can't we use the symmetry of the graph and write $2\pi \int_0^1 (1-y^2)^2 dy$? $\endgroup$ – Jake Nov 22 '14 at 19:56
  • $\begingroup$ Sure you can, either way you'll get the same result. $\endgroup$ – Namaste Nov 22 '14 at 19:58

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