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.