find the volume for a solid rotated about $x=3$

Question

$x=y^2, x = 1-y^2$ rotated about $y=3$. I cannot seem to get the answer, I cannot see any errors

$y^2=1-y^2 \to 2y^2 = 1 \to y = +- \frac{1}{2}$ we have our bounds. Now setting things up. since we are going about $x=3$ I can adjust the equation so that its just like rotating the solid about the y axis.

The larger function with int the bounds is $1-y^2$

$$ \pi \int_{-.5}^{.5} (1-y^2-3)^2 - (y^2-3)^2 dy$$

$$ \pi \int_{-.5}^{.5} (-y^2-2)^2-(y^2-3)^2 dy $$

$(-y^2-2)^2= y^4-2y^2-2y^2+4$

$(y^2-3)^2 = y^4 - 3y^2 - 3y^2 +9$

$ \pi \int_{-.5}^{.5} y^4 - 4y^2 + 4 - y^4 +6y^2 -9$

$2y^2 - 5 \Big\vert_{-.5}^{.5} = \frac{2}{3}y^3 -5y \Big\vert_{-.5}^{.5}$

$\frac{2}{3}(.125)-2.5+ \frac{2}{3}(.125)-2.5$

But this answer is way off and I am not sure why?

Answer 1

Your problems start with $y^2=x=1-y^2\implies y^2=1/2\implies y=\pm\frac1{\sqrt2}$. Then since you are rotating about an axis parallel to the $x$-axis and integrating with respect to $y$, you should be integrating cyclindrical shells. I get $$\begin{align}V&=\int2\pi rh\,dr=\int_{-1/\sqrt2}^{1/\sqrt2}2\pi(3-y)(1-2y^2)dy\\ &=2\pi\int_{-1/\sqrt2}^{1/\sqrt2}(2y^3-6y^2-y+3)dy=2\pi\left[\frac12y^4-2y^3-\frac12y^2+3y\right]_{-1/\sqrt2}^{1/\sqrt2}\\ &=2\pi(2)\left[-\frac2{2\sqrt2}+\frac3{\sqrt2}\right]=4\pi\sqrt2\end{align}$$ Even if the rotation axis were the line $x=3$, observe that $3-(1-y^2)\le3-y^2$ over the interval of integration so you would have inner and outer radii reversed.

Answer 2

The intersection points are wrong: if $2y^2 = 1$, then $y = \pm \frac{1}{\sqrt 2}$. Also, you have a sign error when collecting like terms of $y^2$. We have: \begin{align*} V &= \pi\int_{\frac{-1}{\sqrt 2}}^{\frac{1}{\sqrt 2}} ((3 - y^2)^2 - (3 - (1 - y^2))^2) dy \\ &= \pi\int_{\frac{-1}{\sqrt 2}}^{\frac{1}{\sqrt 2}} ((y^4 - 6y^2 + 9) - (y^4 + 4y^2 + 4)) dy \\ &= \pi\int_{\frac{-1}{\sqrt 2}}^{\frac{1}{\sqrt 2}} (-10y^2 + 5) dy\\ &= -10\pi\int_{0}^{\frac{1}{\sqrt 2}} (2y^2 - 1) dy \\ &= -10\pi\left[ \frac{2y^3}{3} - y \right]_{0}^{\frac{1}{\sqrt 2}} \\ &= \frac{10\sqrt 2}{3}\pi \end{align*}