Question

Sketch the graph and find the volume of the solid obtained by revolving the region bounded by the graphs of x=1, y=x+1 and y=1/sqrt(x^2+1 about the x-axis)

Answer 1

i got an answer of 25pi/12 can someone check if this is right?

Answer 2

first off what shape do you get?

Answer 3

basically draw it out, u know x^2+1 is a parbola i believe you may need to find intersection pts and that will b what ur domain for integral. did u get a disk?

Answer 4

so just find your big area and small and antiderive and plug in for your definite integral

Answer 5

i just use the PiR^2 disk method i think

Answer 6

i graphed it on wolfram http://www.wolframalpha.com/input/?i=graph+x%3D1%2Cy%3Dx%2B1+and+y%3D1%2Fsqrt%28x%5E2%2B1%29

Answer 7

i dont think its a disk its more of a washer, because if u integrate with regards to the x-axis there is a space between the x-axis and your function.

Answer 8

ur going to integrate from 0, 1

Answer 9

thats my integration limits

Answer 10

so use the washer method?

Answer 11

the retarded part is you have 2 functions for your volume LOL typically its 1 lol its a disk though if you want y=1/sqrtx+1

Answer 12

haha yea

Answer 13

i would just integate the disk, i think you did it wiell want me to calcualte?

Answer 14

yes please

Answer 15

i got 0.88

Answer 16

cuz i did the antiderivative of the function because that is your r square it and then integrate, i got that

Answer 17

we do this to calculate the area under the function, but the y=x+1 is throwing me off a bit, if it was one with limits i could do it but this is a bit odd lol

Answer 18

is this your integration? \[\pi \int\limits_{0}^{1}(x+1)^2 - 1/\sqrt(x^2+1)dx\]