The region R in the first quandrant is enclosed by the lines x=0 and y=5 and the curve y=x^2+1. The volume of the solid generated when R is revolved about the y-axisi is

a) 6pi
b) 8pi
c) 34pi/3
d) 16pi
e) 544pi/15

Question

The region R in the first quandrant is enclosed by the lines x=0 and y=5 and the curve y=x^2+1. The volume of the solid generated when R is revolved about the y-axisi is

a) 6pi
b) 8pi
c) 34pi/3
d) 16pi
e) 544pi/15

anonymous · Answer

I am not understanding what im doing wrong but so far my awnser for some reason ends up being 4pi

kainui · Answer

Sure, so show me what you actually integrated or did and I can help you figure out where you went wrong. These are my favorite calculus questions!

anonymous · Answer

Well it being revoled around the y-axis I changed the formula to
x=sqrt(y-1)

so I intergrated this

$$\pi \int\limits_{1}^{5} (\sqrt(y-1)^2dx$$

anonymous · Answer

Oh wait I just relized I forgot to take the anti derivative and I got 8pi

anonymous · Answer

Thats what I did wrong this whole time

anonymous · Answer

so then 

$$\pi([25/2 - 5] - [1/2 -1])=8\pi$$

anonymous · Answer

originally i did

$$\pi([5-1]-[1-1])=4\pi$$

anonymous · Answer

Thanks anyways

kainui · Answer

Haha alright.