Calculus 2 - length of arc and volume questions

[Radam]

https://imgur.com/a/ESsNm

I have a rough idea on how to solve the second one, but I don't know if my method is correct. I used the shell method with the value (y+3) for the radius. But I have no clue how to solve the first question. 

https://imgur.com/a/zJdFw

[studiot]

Are you not aware of this formula?

 

[math]L = \int_a^b {\sqrt {\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)} } dx[/math]

 

[Radam]

2 hours ago, studiot said:

Are you not aware of this formula?

 

L=∫ba(1+(dydx)2)−−−−−−−−−−−√dx

 

Yes I am familiar with it. I'm not sure how I can reduce the square roots, enabling me to integrate. 

[studiot]

Have you tried some algebra?

I make it

[math]4\int_1^2 {\sqrt {1 + \frac{{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)}}{x}} dx} [/math]

 

[Country Boy]

For the first, you are given the graph of y= x⁴+ (1/32)x^-2.  The derivative is y'= 4x³- (1/16)x^-3.  Notice the "3" and "-3" powers.  When you square that, something nice happens: y'²= (4x³- (1/16)x^-3)²= 16x⁶- 2(4x³)(1/16)(x^-3)+ (1/256)x^^-6.  I wrote that middle term out In detail to make clear that the "x" terms cancel leaving 16x⁶- (1/2)+ (1/256)x^-6.  When we add 1 that "-1/2" becomes "+1/2" so that is simply (4x³+ (1/16)x^-3)² , a "perfect square".   sqrt{1+ (dy/dx)²}= 4x³+ (1/16)x^-3.  Integrate that from 1 to 2.

Edited December 7, 2017 by Country Boy