How do you find the volume of the solid obtained by rotating the region bounded by the curves $y = \sqrt{x}$ $y=sqrtx$ and $y = \frac{x}{3}$ $y=x/3$ rotated around the $x = - 1$ $x=-1$ ?

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How do you find the volume of the solid obtained by rotating the region bounded by the curves $y = \sqrt{x}$ $y=sqrtx$ and $y = \frac{x}{3}$ $y=x/3$ rotated around the $x = - 1$ $x=-1$ ?

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Answer 1 · 2015-10-09T16:33:54

See the explanation section, below.

Explanation:

Here is a picture of the region (in blue) and the line $x = - 1$ $x=-1$ .

enter image source here

It shows a representative slice of thickness $d x$ $dx$ . The slice is taken parallel to the axis of rotation, so when we rotate we'll get cylindrical shell.
The volume of the shell is $2 π r h d x$ $2pirhdx$ . The radius, $r$ $r$ , is shown by the dotted red horizontal line. Its length is $x + 1$ $x+1$ .
The height of the shell is the greater $y$ $y$ value minus the lesser $y$ $y$ value. That is: $h = \sqrt{x} - \frac{x}{3}$ $h = sqrtx-x/3$

We also need the point of intersection where $\frac{x}{3} = \sqrt{3}$ $x/3=sqrt3$ . Which happens at $0$ $0$ and at $9$ $9$

The integral we need to evaluate is:

$\int_{0}^{9} 2 π (x + 1) (\sqrt{x} - \frac{x}{3}) d x$ $int_0^9 2 pi (x+1)(sqrtx-x/3) dx$

Expand the integrand to get 4 terms in powers of $x$ $x$ which is straightforward to evaluate.

$2 π \int_{0}^{9} (x + 1) (\sqrt{x} - \frac{x}{3}) d x = \frac{207}{5} π$ $2 pi int_0^9 (x+1)(sqrtx-x/3) dx = 207/5pi$

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How do you find the volume of the solid obtained by rotating the region bounded by the curves $y = \sqrt{x}$ $y=sqrtx$ and $y = \frac{x}{3}$ $y=x/3$ rotated around the $x = - 1$ $x=-1$ ?

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How do you find the volume of the solid obtained by rotating the region bounded by the curves y=sqrtxy=√xy=sqrtx and y=x/3y=x3y=x/3 rotated around the x=-1x=−1x=-1?

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How do you find the volume of the solid obtained by rotating the region bounded by the curves $y = \sqrt{x}$ $y=sqrtx$ and $y = \frac{x}{3}$ $y=x/3$ rotated around the $x = - 1$ $x=-1$ ?