check answer @solomonzelman

Question

studygurl14 · Answer

attachment

solomonzelman · Answer

Yes, you are right

studygurl14 · Answer

thank you!

solomonzelman · Answer

what if it was revolved around y-axis, then?

studygurl14 · Answer

umm...I'd have to solve for x and use the limits 8 and 0

solomonzelman · Answer

Limits are y=1 and y=8.

$\color{#000000}{    \displaystyle  y=(x^2+1)^3         }$

$\color{#000000}{    \displaystyle\sqrt[3]{y}=x^2+1        }$

$\color{#000000}{    \displaystyle\sqrt[3]{y}-1=x^2        }$

$\color{#000000}{    \displaystyle\sqrt{\sqrt[3]{y}-1}=\sqrt{x^2}        }$

$\color{#000000}{    \displaystyle\sqrt{\sqrt[3]{y}-1}=|x|        }$

$\color{#000000}{    \displaystyle \pm\sqrt{\sqrt[3]{y}-1}=x        }$

the negative goes off since the curve is entirely positive.

$\color{#000000}{    \displaystyle\sqrt{\sqrt[3]{y}-1}=x        }$

solomonzelman · Answer

$\color{#000000}{    \displaystyle\pi \int_1^8(\sqrt[3]{y}-1)dy        }$

solomonzelman · Answer

would not be difficult if revolved around either of the axes.

solomonzelman · Answer

Note, that I mean entirely positive for our purposes, because we are considering the 1st-quadrant part of the function as a boundary.

solomonzelman · Answer

In any case, won't overwhe$\mathcal{L}\{m\}$