# Thread: Rotational solids? Drawing a blank...

1. ## Rotational solids? Drawing a blank...

Just needed to check something with you math experts ;o)

1. Find the volume of the solid obtained by rotating the region bounded by $y=\sqrt[6]{x}$ and $y=x$, rotated about the line $y=1$

Is it

$\int_0^1\pi{(1-\sqrt[6]{x})-(1-x)}$ ??

2. Same premise as the previous problem, but with the following data:
region bounded by $y=x^4$ and $x=y^4$, rotated about $x=-1$

Now, this really got me. Am I supposed to put it all in terms of $x$, so that the regions would be $y=x^4$ and $y=\sqrt[4]{x}$ ?? I'm having a little trouble graphing these... but I think the lower limit is zero...?

Lastly,
3. When I'm told to use the method of cylindrical shells to find a volume, is that the following equation:

$\int_a^b{2\pi}({r})({h})dx$ ??

2. Originally Posted by Kimberly
Just needed to check something with you math experts ;o)

1. Find the volume of the solid obtained by rotating the region bounded by $y=\sqrt[6]{x}$ and $y=x$, rotated about the line $y=1$

Is it

$\int_0^1\pi{(1-\sqrt[6]{x})-(1-x)}$ ??

2. Same premise as the previous problem, but with the following data:
region bounded by $y=x^4$ and $x=y^4$, rotated about $x=-1$

Now, this really got me. Am I supposed to put it all in terms of $x$, so that the regions would be $y=x^4$ and $y=\sqrt[4]{x}$ ?? I'm having a little trouble graphing these... but I think the lower limit is zero...?

Lastly,
3. When I'm told to use the method of cylindrical shells to find a volume, is that the following equation:

$\int_a^b{2\pi}({r})({h})dx$ ??

#1 ...

disks w/r to x

$\displaystyle V = \pi \int_0^1 (1-x)^2 - (1-\sqrt[6]{x})^2 \, dx
$

shells w/r to y

$\displaystyle V = 2\pi \int_0^1 (1-y)(y-y^6) \, dy$

#2 might be easier using shells w/r to x

$\displaystyle V = 2\pi \int_0^1 (x+1)(\sqrt[4]{x} - x^4) \, dx$

#3 $\displaystyle V = 2\pi \int_a^b r(x) \cdot h(x) \, dx$