- #1
lucidlobster
- 5
- 0
Find the area of the given function, rotated about the y axis
The Area below the line y=2 and above y=sin(x) from 0-\pi
I did this rotated around the X AXIS no problem,
I found the area of the disk to be \pi(4-sin^{2}(x))
The volume around the X AXIS to be\frac{7\pi^{2}}{2}
For the Y axis I have used this:\int^{\pi}_{0} 2\pi xf(x)dx
Plugging in:
\int^{\pi}_{0} 2\pi x(2-sin(x))dx
I solved this to be 2\pi^{3}-2\pi^{2}
Does this look right?
The Area below the line y=2 and above y=sin(x) from 0-\pi
I did this rotated around the X AXIS no problem,
I found the area of the disk to be \pi(4-sin^{2}(x))
The volume around the X AXIS to be\frac{7\pi^{2}}{2}
The Attempt at a Solution
For the Y axis I have used this:\int^{\pi}_{0} 2\pi xf(x)dx
Plugging in:
\int^{\pi}_{0} 2\pi x(2-sin(x))dx
I solved this to be 2\pi^{3}-2\pi^{2}
Does this look right?
