At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

hmmm....we just did one like like this earlier :)

hello :)) ya its the exact same one. im a little stuck

y = 5; means that we take the functions and -5 to get them to y=0

5x-5 and sqrt(x)-5 rotated about the x axis

yeah, both of those are squared right. then i take the anti derivative of it correct ?

5x = sqrt(x)
25x^2 = x
25x^2 -x = 0
x(25x - 1) = 0
x = 0 and x = 1/25 right?

yup

then our bounds of integration are 0 to 1/25

im a little confused now because i thought our bounds were from 0 to 5 from before

before you had 5sqrt(x). Did you typo it here?

no thats correct but when we first graphed the two equations, werent the bounds from 0 to 5 ?

when does 5x = sqrt(x)
wehn x = 0 and x = 1/25
5/25 = 1/5 and sqrt(1/25) = 1/5 right?

yup

we arent double integrating this are we?

ohh nooo we arent. sorry sorry. i was mixing it up with something else

yes

pi [S] [5x-5]^2 - [sqrt(x)-5]^2 dx ; [0,1/25]

25x^3 51x^2 20sqrt(x^3)
------ - ------- + ----------
3 2 3

and then we plug in the boundaries, and subtract them right?

yes; but the easy thing here is that 0 = 0 so we only determine 1/25 for our answer

ah. gotcha. thank you so much for your help. made it a lot clearer for me. thank you :)

youre welcome :)

dot worry if the volume is negative; just take the absolute value of it then :)