volume of a solid: region bdd by y=4, y=3sin(x)+1, -pi/2<=x<=3pi/2, abt y=4

[bommar57]

Find the volume of the solid generated by rotating the region of the x,y-plane between the line y = 4, the curve y = 3 sin(x) + 1, for \(\displaystyle -\pi/2\, \leq\, x\, \leq\, 3\pi/2\) about the line y = 4.

I am having trouble with this one, I think it would be the integral from -pi/2 to 3pi/2 of (4-3sinx+1)^2 and when i integrate that it keeps telling me I am incorrect and gives me the hint of using a double angle formula but i dont have the slightest clue on how to do that. My main question I guess is if my integral is set up correctly? thanks

 

[stapel]

Find the volume of the solid generated by rotating the region of the x,y-plane between the line y = 4, the curve y = 3 sin(x) + 1, for \(\displaystyle -\pi/2\, \leq\, x\, \leq\, 3\pi/2\) about the line y = 4.

I am having trouble with this one, I think it would be the integral from -pi/2 to 3pi/2 of (4-3sinx+1)^2 and when i integrate that it keeps telling me I am incorrect...

What is the "it" that is "telling" you this? Also, why are you not, for instance, simplifying the 4 and the 1 into a 5?

When you reply, please show your work so far. Thank you!

 

[pka]

Find the volume of the solid generated by rotating the region of the x,y-plane between the line y = 4, the curve y = 3 sin(x) + 1, for \(\displaystyle -\pi/2\, \leq\, x\, \leq\, 3\pi/2\) about the line y = 4.

I am having trouble with this one, I think it would be the integral from -pi/2 to 3pi/2 of (4-3sinx+1)^2 and when i integrate that it keeps telling me I am incorrect and gives me the hint of using a double angle formula but i dont have the slightest clue on how to do that. My main question I guess is if my integral is set up correctly? thanks

You missed the sign.Where is the \(\displaystyle \pi~?\) as in \(\displaystyle \pi r^2\)

Here is the calculation.