# Thread: Volume by slices and shells

1. ## Volume by slices and shells

Question: Let V be the volume of the solid obtained by rotating about the y-axis the region bounded by $\displaystyle y=\sqrt{x}$ and $\displaystyle y=x^2$. Find V both by slicing and by cylindrical shells. In both cases, draw a diagram to explain your method.

My work:
-------------------
a) Volume by slices.
Since this is rotated about the y-axis, I solved in terms of x:
$\displaystyle y=\sqrt{x} ~~~~\Rightarrow ~~~~x=y^2$
$\displaystyle y=x^2 ~~~~~\Rightarrow ~~~~x = \sqrt{y}$

Then I said the cross-sectional areal will be $\displaystyle \pi x^2$ since x will be the radius.

So I plugged my equations into it:
$\displaystyle Area_1 = \pi y$
$\displaystyle Area_2 = \pi y^2$

Since they intersect at the points (0,0) and (1,1) Volume will be from y=0 to y=1:
$\displaystyle V=\int_0^1 (Area_1-Area_2)~dy$

$\displaystyle V=\pi \int_0^1 (y-y^2)~dy$

$\displaystyle V=\pi (\frac 12y^2-\frac 13y^3)_0^1$

$\displaystyle V=\pi (\frac 12-\frac 13)$

$\displaystyle V=\frac 16\pi$

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b) volume by shells
Using the initial equations, solved for y, volume will be the volume of $\displaystyle y=x^2$ - volume of $\displaystyle y=\sqrt{x}$. from x=0 to x=1.

So volume = $\displaystyle 2\pi \int_0^1x(x^2)~dx - 2\pi \int_0^1 x(\sqrt{x})~dx$

$\displaystyle V = 2\pi \int_0^1 (x^3-x^{3/2})~dx$

$\displaystyle V = 2\pi (\frac 14x^4-\frac 25x^{5/2})_0^1$

$\displaystyle V = 2\pi (\frac 14-\frac 25)$

$\displaystyle V = 2\pi (-\frac 3{20})$

$\displaystyle V = -\frac{6}{20}\pi$

So either one or both of my ways are wrong >.<

2. Originally Posted by angel.white
Question: Let V be the volume of the solid obtained by rotating about the y-axis the region bounded by $\displaystyle y=\sqrt{x}$ and $\displaystyle y=x^2$. Find V both by slicing and by cylindrical shells. In both cases, draw a diagram to explain your method.

My work:
-------------------
a) Volume by slices.
Since this is rotated about the y-axis, I solved in terms of x:
$\displaystyle y=\sqrt{x} ~~~~\Rightarrow ~~~~x=y^2$
$\displaystyle y=x^2 ~~~~~\Rightarrow ~~~~x = \sqrt{y}$

Then I said the cross-sectional areal will be $\displaystyle \pi x^2$ since x will be the radius.

So I plugged my equations into it:
$\displaystyle Area_1 = \pi y$
$\displaystyle Area_2 = \pi y^2$

Since they intersect at the points (0,0) and (1,1) Volume will be from y=0 to y=1:
$\displaystyle V=\int_0^1 (Area_1-Area_2)~dy$

$\displaystyle V=\pi \int_0^1 (y-y^2)~dy$

$\displaystyle V=\pi (\frac 12y^2-\frac 13y^3)_0^1$

$\displaystyle V=\pi (\frac 12-\frac 13)$

$\displaystyle V=\frac 16\pi$

[snip]
The formula is $\displaystyle V = \pi \int_{y=a}^{y=b} x^2 \, dy$.

So:

$\displaystyle V_1 = \pi \int_{y=0}^{y=1} y \, dy$.

Note: $\displaystyle y = x^2 \Rightarrow x^2 = y$.

$\displaystyle V_2 = \pi \int_{y=0}^{y=1} y^4 \, dy$.

Note: $\displaystyle y = \sqrt{x} \Rightarrow x = y^2 \Rightarrow x^2 = y^4$.

$\displaystyle V = V_1 - V_2 = \frac{3\, \pi}{10}$ cubic units.

3. Originally Posted by angel.white
[snip]
b) volume by shells
Using the initial equations, solved for y, volume will be the volume of $\displaystyle y=x^2$ - volume of $\displaystyle y=\sqrt{x}$. from x=0 to x=1.

Mr F says: Other way around! That is, volume of $\displaystyle {\color{red}y=\sqrt{x}}$ - volume of $\displaystyle {\color{red}y=x^2}$.

Note: From a graph of the region, it's clear that the height of the shells is $\displaystyle {\color{red} \sqrt{x} - x^2}$ since $\displaystyle {\color{red}y=\sqrt{x}}$ is above $\displaystyle {\color{red}y=x^2}$ over the interval 0 < x < 1.

So volume = $\displaystyle 2\pi \int_0^1x(x^2)~dx - 2\pi \int_0^1 x(\sqrt{x})~dx$

$\displaystyle V = 2\pi \int_0^1 (x^3-x^{3/2})~dx$

$\displaystyle V = 2\pi (\frac 14x^4-\frac 25x^{5/2})_0^1$

$\displaystyle V = 2\pi (\frac 14-\frac 25)$

$\displaystyle V = 2\pi (-\frac 3{20})$

$\displaystyle V = -\frac{6}{20}\pi$
[snip]
Very close. The answer is $\displaystyle \frac{6}{20} \, \pi = \frac{3 \, \pi}{10}$ cubic units, in agreement with the answer found using the slices method.