Question

help please

number 6

Answer 1

the curve y=x^2 , 0</ x </ 1 is rotated about the y axis. find the area of the resulting surface

Answer 2

You should be able to do this... start with a diagram.

Answer 3

this is where i left of .

Answer 4

yeah looks alright so far.. you want to integrate over x=0 to x=1

y = x^2
y ' = 2x dx

\[\large 2 \pi \int\limits_{0}^{1}x \sqrt{1+4x^2}dx\]

Answer 5

Sub

u = 1+ 4x^2
du = 8x dx

Answer 6

yes, also make sure to change the bounds, x from 0 to 1

u = 1 + 4x^2

becomes u from 1 to 5

Answer 7

so then i got this

\[\frac{ \pi }{ 6 }(1+4x^2)^3/2 ...from 0\to1\]

Answer 8

\[\large 2\pi \int\limits_{0}^{1}x \sqrt{1+4x^2}dx\]

if u = 1+4x^2
du = 8x dx
so
x*dx = 1/8 du

\[\large 2\pi*\int\limits_{1}^{5}\sqrt{u}*\frac{ 1 }{ 8 }du\]

Answer 9

notice also it is from u=1 to u=5 now

Answer 10

oh riightt

Answer 11

\[= \large \frac{ 2\pi }{ 8 }*\frac{ 2 }{ 3 }*u^{\frac{ 3 }{ 2} }\]
from u=1 to u=5