Question

This problem seems so easy but idk what I'm doing
(Calculus)

Answer 1

This is me putting my rotation line around x=5, and so it creates a sort of washer shape I assume?
Where we take distance from top function to axis of rotation as our big radius and distance from bottom function to axis of rotation as our little radius?

Answer 2

So then
\[V=\int\limits_{-1}^{1}(\pi*(x^2+5)^2-\pi*(4)^2)dx\]
???
I used -1 and 1 as x-bounds and formula for a washer shape object is \[\pi*R^2-\pi*r^2\]

Answer 3

distance from x^2 to line 5 I represented as x^2+5 and distance from the line y=5 to y=1 I have as 4

Answer 4

@vocaloid

Answer 5

Unless you are making an effort to help, Do not reply

Answer 6

@AZ i understand this but i dont want to give you the wrong steps I have a weird way of calculating things in my head.....:)

Answer 7

@kxylaaaaa i know your very reliable with calc help out

Answer 8

\[volume=\pi \int\limits_{-1}^{1}[(5-x^2)^2-4^2]dx\]
\[=\pi \int\limits_{-1}^{1}[25+x^4-10x^2-16]dx\]
if we change x to -x function remains the same.
\[volume=2 \pi \int\limits_{0}^{1}(9+x^4-10x^2)dx\]
\[=2 \pi (9x+\frac{ x^5 }{ 5 }-\frac{ 10x^3 }{ 3 })|x ~from 0 \rightarrow 2\]
\[=2 \pi(9+\frac{ 1 }{ 5 }-\frac{ 10 }{ 3 })\]
\[=2 \pi(9-\frac{ 47 }{ 15 })\]
\[=2 \pi (\frac{ 135-47 }{ 15 })\]
\[=2\pi \times \frac{ 88 }{ 15 }\]
\[=\frac{ 176 \pi }{ 15 }\]
\[\approx36.8613538021\]

Answer 9

So 5-x^2 and not (x^2)+5 for distance, true, the distance at most can only be 5 so (x^2)+5 isn't correct, thanks Surjithayer

Answer 10

yw

Answer 11

correction x varies from 0 to 1 and not 0 to 2