Question

Find the exact area of the surface obtained by rotating the curve about the x-axis.

Answer 1

\[x = \frac{ 1 }{ 3 }(y^2+2)^{3/2}, 4 \le y \le 5\]

Answer 2

you use
\[\int\limits 2 \pi y ds \\ \text{ where } ds=\sqrt{1+(y')^2} dx , a \le x \le b \\ \text{ or } \\ \text{ use } ds=\sqrt{1+(x')^2} dy , c \le y \le d \\ \\ \text{ by the way when I say } y' \text{ I mean } \frac{dy}{dx} \\ \text{ and when I said } x' \text{ I mean } \frac{dx}{dy}\]

Answer 3

I think I prefer the second ds

Answer 4

\[x'=\frac{ 1 }{ 3 }[\frac{ 3 }{ 2 }[y^2+2]^{1/2}*(2y)]\] Did I find the x' correctly?

Answer 5

looks good
can be simplified a bit

Answer 6

\[x ' = y \sqrt{y^2+2}\]

Answer 7

\[x'=y(y^2+1)^\frac{1}{2}\]
yep yep

Answer 8

and (x')^2=?

Answer 9

\[y^2(y^2+2)\]

Answer 10

you probably got through all of this but got scared of the square root part
I will give you a hint 1+(x')^2 ends up being a perfect square

Answer 11

good you used your x'
I made a type-o in my above :p

\[x'=y(y^2+2)^\frac{1}{2} \\ (x')^2=y^2(y^2+2) \\ (x')^2=y^4+2y^2 \\ (x')^2+1=y^4+2y^2+1\]

Answer 12

\[(x')^2+1=(y^2+1)^2 \]

Answer 13

\[\sqrt{(x')^2+1}=\sqrt{(y^2+1)^2}=y^2+1 \text{ since } y^2+1 \text{ is always positve }\]

Answer 14

\[S = 2\pi \int\limits_{4}^{5}y \sqrt{1+y^4+2y^2}dy\]

Answer 15

\[S = 2\pi \int\limits_{4}^{5}y(y^2+1)dy=2\pi \int\limits_{4}^{5}(y^3+y)dy\]
Like that?

Answer 16

yes

Answer 17

Okay, i got the problem now, just needed help setting the square root up and using x' instead of y'. Thanks :D

Answer 18

np