Find the volume of the solid generated by revolving the area.

[bob29]

Homework Statement

Find the volume of the solid generated by revolving the area bounded by
y = sec[x], y= [tex]\sqrt{2}[/tex]
- [tex]\pi[/tex]/4 [tex]\leq[/tex] x [tex]\leq[/tex] [tex]\pi[/tex]/4 about the x-axis

Or
[PLAIN]http://img838.imageshack.us/img838/1552/mathprobq6.jpg

Homework Equations

a = lower limit = --[tex]\pi[/tex]/4
b = upper limit = [tex]\pi[/tex]/4
V = [tex]\int[/tex] a->b R² - r² dx

The Attempt at a Solution

V = [tex]\int[/tex] a->b R² - r² dx
V = [tex]\pi[/tex] [tex]\sqrt{2}[/tex]² - [tex]\pi[/tex]*sec[x] = a->b([tex]\pi[/tex]*2x) - ([tex]\pi[/tex]*[tan[x]])a->b

Note: Ans = [tex]\pi[/tex][[tex]\pi[/tex] - 2]

 

[Mark44]

Homework Statement

Find the volume of the solid generated by revolving the area bounded by
y = sec[x], y= [tex]\sqrt{2}[/tex]
- [tex]\pi[/tex]/4 [tex]\leq[/tex] x [tex]\leq[/tex] [tex]\pi[/tex]/4 about the x-axis

Or
[PLAIN]http://img838.imageshack.us/img838/1552/mathprobq6.jpg

Homework Equations

a = lower limit = --[tex]\pi[/tex]/4
b = upper limit = [tex]\pi[/tex]/4
V = [tex]\int[/tex] a->b R² - r² dx

The Attempt at a Solution

V = [tex]\int[/tex] a->b R² - r² dx
V = [tex]\pi[/tex] [tex]\sqrt{2}[/tex]² - [tex]\pi[/tex]*sec[x] = a->b([tex]\pi[/tex]*2x) - ([tex]\pi[/tex]*[tan[x]])a->b

Note: Ans = [tex]\pi[/tex][[tex]\pi[/tex] - 2]

What's your question? You have the integral set up correctly (but horribly formatted). The only things missing are the limits of integration.

Here's your integral, formatted a little more nicely. Click the integral to see what I did.
[tex]V = \pi \int_a^b \left( 2 - sec^2(x) \right)dx[/tex]

 

[bob29]

Thanks anyway, yep not familiar with the formatting.
But problem is solved.