Nash Frank

2022-07-16

Finding the volume by Shell method: $y={x}^{2},y=2-{x}^{2}$, $y={x}^{2},y=2-{x}^{2}$ about the line $x=1$.
what I get from this after graphing is:
$2\pi \int \left(1-x\right)\left(2-2{x}^{2}\right)dx$
which becomes: $2\pi \int \left(2-2{x}^{2}-2x+2{x}^{3}\right)dx$
integrating that I get:
$2\pi \left[2x-\frac{2}{3}{x}^{3}-{x}^{2}+\frac{1}{2}{x}^{4}\right]$ from 0 to 1
My answer is $\frac{5}{3}\pi$ but my book says the answer is: $\frac{16}{3}\pi$
Could someone tell me where I went wrong? Was it the upper lower bounds?

Monastro3n

Expert

Explanation:
The limits of the integration is given by where the lines cross each other:
${x}^{2}=2-{x}^{2}⟺x=±1$
so the limits shold be from -1 to 1.

Expert

Explanation:
The region in the xy-plane bounded by the graphs $y={x}^{2}$ and $y=2-{x}^{2}$ is symmetric about the y-axis, so the projection of this region onto the x-axis should be a symmetric interval of the form [-a,a]. In this case the points of intersection of the graphs have x-coordinates $x=±1$. Thus, your limits of integration should really be from $x=-1$ to $x=1$, not 0 to 1.

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