Posted by **Naseba** on Wednesday, April 4, 2012 at 10:59pm.

Let R be the region in the first quadrant that is enclosed by the graph of y = tanx, the x-axis, and the line x = π/3

h. Find the area of R

i. Find the volume of the solid formed by revolving R about the x-axis

- calculus -
**Steve**, Thursday, April 5, 2012 at 11:09am
what's the problem?

A = ∫[0,pi/3] tanx dx

integrating tanx is easy, since

tanx = sinx/cosx = -d(cosx)/cosx

∫tanx dx = -ln(cosx)

A = -ln(cosx) [0,pi/3]

= -ln(cos pi/3) + ln(cos 0)

= -ln(1/2) + ln(1)

= ln2

v = ∫[0,pi/3] pi*r^2 dx

= pi*∫[0,pi/3] tan^2x dx

= pi*∫[0,pi/3](sec^2x - 1) dx

= pi*(tanx - x) [0,pi/3]

. . .

## Answer this Question

## Related Questions

- Calculus 1 - Let f be the function given by f(x)=3sqrt(x-2). A) On the axes ...
- calculus - 1. Let R be the region in the first quadrant enclosed by the graphs ...
- calculus - Which has more area, the region in the first quadrant enclosed by the...
- AP Calculus - Let R be the first quadrant region enclosed by the graph of y= 2e...
- calculus - Find the area of the region in the first quadrant enclosed by the ...
- Calculus - I am doing the AP calculus review, these are the questions I have no ...
- Calculus - This problem set is ridiculously hard. I know how to find the volume ...
- Calculus - Let f and g be the functions given by f(x)=1+sin(2x) and g(x)=e^(x/2...
- Calculus - Let f be the function given by f(x)=(x^3)/4 - (x^2)/3 - x/2 + 3cosx. ...
- Calculus - Let R be the region in the first quadrant under the graph of y=1/sqrt...

More Related Questions