How do you find the volume of the region bounded by the graph of $y = x^{2} + 1$ $y = x^2+1$ for x is [1,2] rotated around the x axis?

Question

How do you find the volume of the region bounded by the graph of $y = x^{2} + 1$ $y = x^2+1$ for x is [1,2] rotated around the x axis?

1 Answer

Impact of this question

1712 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-12T17:40:24

See below.

Explanation:

The region is in blue. The rotation is shown by the red circular arrow.

A representative slice has been taken perpendicular to the axis or rotation to use the method of discs. The slice has black borders.

enter image source here

The volume of a representative slice is

$π r^{2} \cdot thickness$ $pir^2* "thickness"$

In general, the radius $r$ $r$ is the longer distance minus the shorter. In this case the shorter distance to the axis of rotation is $0$ $0$ so we have

$r = x^{2} + 1$ $r = x^2+1$

The thickness is a differential. It is the thin side of the slice. IN this case $thickness = d x$ $"thickness" = dx$

The values of $x$ $x$ go from $1$ $1$ to $2$ $2$ .

The volume we seek is

$V = \int_{1}^{2} π {(x^{2} + 1)}^{2} d x$ $V = int_1^2 pi (x^2+1)^2 dx$

$= π \int_{1}^{2} (x^{4} + 2 x^{2} + 1) d x$ $= pi int_1^2 (x^4+2x^2+1) dx$

$= \frac{178}{15} π$ $= 178/15 pi$

Science

Math

Humanities

... and beyond

How do you find the volume of the region bounded by the graph of $y = x^{2} + 1$ $y = x^2+1$ for x is [1,2] rotated around the x axis?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the volume of the region bounded by the graph of y=x2+1y = x^2+1 for x is [1,2] rotated around the x axis?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the volume of the region bounded by the graph of $y = x^{2} + 1$ $y = x^2+1$ for x is [1,2] rotated around the x axis?