Question

See the post for my question

Answer 1

A curve is defined as\[y=\frac{ 1 }{ 3 }x ^{3}+1\]
The surface area generated when this curve is rotated throughgh one complete revolution about the x-axis us denoted by S
Sow that \[S=\frac{ 1 }{ 9 }\pi \left( 18s+2\sqrt{2} -1\right)\]
Where s is defined as\[\int\limits_{0}^{1} \sqrt{1+x ^{4}}dx \]

Answer 2

@hartnn

Answer 3

do you want to find the value of s???

Answer 4

@sauravshakya It is mentioned in the question " Do not attempt to evaluate s or S "

Answer 5

are there bounds on x ?

Answer 6

Yes, sure. If you look at the boundaries for s, you will notice it's 0 to 1

Answer 7

the surface area is

2pi y ds
or
2 pi (x^3 /3 + 1 ) ds

now you need ds
where s is the arc length

Answer 8

the arc length is given as s

Answer 9

the easy part is
\[2 \pi \int\limits_{0}^{1}ds + 2 \pi \int\limits_{0}^{1} \frac{x^3}{3} ds= 2 \pi s + 2 \pi \int\limits_{0}^{1} \frac{x^3}{3} ds\]
or
\[ \frac{1}{9} \pi \left( 18s + 18 \int_0^1 \frac{x^3}{3} ds \right) \]

Answer 10

now we need to show
\[ 6 \int_0^1 x^3 ds = 2 \sqrt{2} -1 \]

Answer 11

ds= sqrt( 1+ x^4 ) dx , is that right?

Answer 12

oh! so use substitution! Get it!

Answer 13

yes, so it does not look too bad

Answer 14

u= 1+x^4 du = 4 x^3 dx

\[ \frac{6}{4} \int_0^1 u^{\frac{1}{2}} du \\
= \frac{3}{2} \frac{2}{3} u^{\frac{3}{2}}
\]

Answer 15

thx! you are the rescue!