Question

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

Answer 1

x=1/3(y^2+2)^3/2, 1<y<3

Answer 2

hello

Answer 3

hi

Answer 4

could you help me with this problem

Answer 5

the form of an integral rotating about x axis is
$$\huge\int 2\pi~ y ~ ds$$

Answer 6

okay so how do i get y and ds

Answer 7

you have two options

$$\huge\int 2\pi~ f(x) ~ \sqrt{1+(dy/dx)^2}dx$$

or
$$\huge\int 2\pi~ y ~ \sqrt{1+(dx/dy)^2}dy$$

Answer 8

can we do the second one?

Answer 9

yes

Answer 10

or whichever is the easiest

Answer 11

that will be easiest since we were given x = g(y)

Answer 12

okay

Answer 13

so first find dx/dy

Answer 14

i got 1/2 * (y^2+2)^1/2 * 2y

Answer 15

$$x=1/3(y^2+2)^{3/2} \\ \frac{dx}{dy} = 1/3 *(3/2)(y^2+2)^{1/2} * 2y $$

Answer 16

did i copy x(y) correctly?

Answer 17

yes that right

Answer 18

ok i got same as you

Answer 19

$$\huge \int_{1}^{3} 2\pi y \sqrt{ 1+(y*(y^2+2)^{1/2} )^2}~dy $$

Answer 20

okay

Answer 21

can we multiply the y in

Answer 22

okay but what happened to the 1/2?

Answer 23

$$ \Large x=1/3(y^2+2)^{3/2} \\
\Large \frac{dx}{dy} = 1/3 *(3/2)(y^2+2)^{1/2} * 2y \\
\Large \frac{dx}{dy}= 1/3 * 3/2 * 2 y (y^2+2)^{1/2} \\

\Large \frac{dx}{dy}= y (y^2+2)^{1/2}

$$

Answer 24

alright got that

Answer 25

so now all we need is the y right?

Answer 26

$$
\Large \int_{1}^{3} 2\pi y \sqrt{ 1+(y*(y^2+2)^{1/2} )^2}~dy
\\ \Large \int_{1}^{3} 2\pi y \sqrt{ 1+y^2(y^2+2)}~dy
\\\Large \int_{1}^{3} 2\pi y \sqrt{ y^4+2y^2+1}~dy
\\\Large \int_{1}^{3} 2\pi y \sqrt{ (y^2)^2+2(y^2)+1}~dy

\\ \Large \int_{1}^{3} 2\pi y \sqrt{ (y^2+1)(y^2+1)}~dy
\\ \Large \int_{1}^{3} 2\pi y \sqrt{ (y^2+1)^2}~dy
\\ \Large \int_{1}^{3} 2\pi y ~ (y^2+1)~dy

\\ \Large 2\pi \int_{1}^{3} (y^3+y)~dy


$$

Answer 27

wow thats alot of y's

Answer 28

1/4 y^4 + 1/2 y^2

Answer 29

do you have to factor out the 1/4 and 1/2?

Answer 30

you shouldn't

Answer 31

okay

Answer 32

thank you

Answer 33

do you do the same steps if the curve rotates about the y-axis?

Answer 34

yes

Answer 35

okay thanks

Answer 36

$$\\ \Large 2\pi \int_{1}^{3} (y^3+y)~dy= 2\pi (\frac{y^4}{4}+\frac{y^3}{3})\LARGE|_1^3
$$

Answer 37

@perl one quick question what did you do to the 2 and the 2 exponent on the fourth line?

Answer 38

can you point to it , or draw it