Nash Frank

Answered

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:

$2\pi \int (1-x)(2-2{x}^{2})dx$

which becomes: $2\pi \int (2-2{x}^{2}-2x+2{x}^{3})dx$

integrating that I get:

$2\pi [2x-\frac{2}{3}{x}^{3}-{x}^{2}+\frac{1}{2}{x}^{4}]$ 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?

what I get from this after graphing is:

$2\pi \int (1-x)(2-2{x}^{2})dx$

which becomes: $2\pi \int (2-2{x}^{2}-2x+2{x}^{3})dx$

integrating that I get:

$2\pi [2x-\frac{2}{3}{x}^{3}-{x}^{2}+\frac{1}{2}{x}^{4}]$ 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?

Answer & Explanation

Monastro3n

Expert

2022-07-17Added 15 answers

Explanation:

The limits of the integration is given by where the lines cross each other:

${x}^{2}=2-{x}^{2}\u27fax=\pm 1$

so the limits shold be from -1 to 1.

The limits of the integration is given by where the lines cross each other:

${x}^{2}=2-{x}^{2}\u27fax=\pm 1$

so the limits shold be from -1 to 1.

Haley Madden

Expert

2022-07-18Added 7 answers

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 $x=\pm 1$. Thus, your limits of integration should really be from $x=-1$ to $x=1$, not 0 to 1.

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 $x=\pm 1$. Thus, your limits of integration should really be from $x=-1$ to $x=1$, not 0 to 1.

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