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= \sqrt{2}
- \pi/4 \leq x \leq \pi/4 about the x-axis

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

Homework Equations

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

The Attempt at a Solution

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

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

 

[Mark44]

Homework Statement

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

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

Homework Equations

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

The Attempt at a Solution

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

Note: Ans = \pi[\pi - 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.
V = \pi \int_a^b \left( 2 - sec^2(x) \right)dx

 

[bob29]

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