Question

find the area of the surface generated by revolving the curve on the interval 0 11 years ago

Answer 1

x = t and y= 7t

Answer 2

@perl help me please

Answer 3

what is the given axis

Answer 4

it doesnt tell me

Answer 5

just says about the given axis....

Answer 6

when that happens can i use either axis y or x?

Answer 7

your good @perl

Answer 8

yeah he is one of the best

Answer 9

we might get difference answers if you rotate about the y axis versus rotating about x axis

Answer 10

my teacher gave us this problem i guess he forgot to pick an axis

Answer 11

per i wish i cod be mod or ambassdor

Answer 12

$$\Large{
\text{rotation about x axis: }\int2\pi y ds :\\
\text{rotation about y axis: }\int 2 \pi x ds :
}
$$

Answer 13

okay and i'll get the same answer from both of them

Answer 14

$$\Large{
\text{rotation about x axis: }\\
\int_{0}^{3}2\pi y \sqrt{ \left( \frac{dx}{dt} \right)^2+ \left( \frac{dy}{dt} \right) ^2} ~dt \\


}
$$

Answer 15

$$\Large{x = t ,y= 7t\\
\text{rotation about x axis: }\\
\int_{t_1}^{t_2}2\pi y \sqrt{ \left( \frac{dx}{dt} \right)^2+ \left( \frac{dy}{dt} \right) ^2} ~dt
\\
=\int_{0}^{3}2\pi \cdot 7t \sqrt{ \left( 1 \right)^2+ \left( 7 \right) ^2} ~dt
\\

\\


}
$$

Answer 16

dx = 1 and dy = 7 right?

Answer 17

okay and i dont need to change the integrals for this problem correct?

Answer 18

the t2 and t1 remain 3 and 0 right

Answer 19

right

Answer 20

surface of area of revolution does change

Answer 21

what do you mean

Answer 22

$$


\Large{x = t ,y= 7t\\
\text{rotation about x axis: }\\
\int_{t_1}^{t_2}2\pi y \sqrt{ \left( \frac{dx}{dt} \right)^2+ \left( \frac{dy}{dt} \right) ^2} ~dt
\\
=\int_{0}^{3}2\pi \cdot 7t \sqrt{ \left( 1 \right)^2+ \left( 7 \right) ^2} ~dt
\\
=\sqrt{50}\int_{0}^{3}2\pi \cdot 7t ~dt
\\
=315 \pi \sqrt 2
\\ \therefore \\
\text{rotation about y axis: }\\
\int_{t_1}^{t_2}2\pi x \sqrt{ \left( \frac{dx}{dt} \right)^2+ \left( \frac{dy}{dt} \right) ^2} ~dt
\\
=\int_{0}^{3}2\pi \cdot t \sqrt{ \left( 1 \right)^2+ \left( 7 \right) ^2} ~dt
\\
=\sqrt{50}\int_{0}^{3}2\pi \cdot t ~dt
\\
=45 \pi \sqrt 2
\\

}
$$

Answer 23

okay so the answers are not the same

Answer 24

if i ask one more question on here would you help me it'll be the last one for the night

Answer 25

i just want to know if my work is right

Answer 26

could you?

Answer 27

ok

Answer 28

alright thank you

Answer 29

i am going to draw it out

Answer 30

this is problem find the arc length of the curve on the interval 0<t<6

Answer 31

x = sqrt(t) and y = 9t-6

Answer 32

for dx/dt = 1/2sqrt(t) and dy/dt = 9