area/volume of region enclosed by y=tan x, the x-axis, and the line x=pi/3?

Question

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

a) find the area of R

b)find the volume of the solid formed by revolving R about the x-axis.

Answer 1

Ok, the area enclosed will be :

integrate(tanx) from 0 to pi/3

integrate(sinx / cosx)dx

-integrate(dcosx / cosx)dx

-lnx(cosx)dx from 0 to pi/3

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

-ln(1/2) = area of R.

Ln(2) = area of R

b) the volume I think you can get it using Pappus Gulding law :

volume = 2pi*area*pi/6

volume = 2pi*ln(1/2)*pi/6

volume = pi^2 / 2*ln(1/2)

Hope that helps

Answer 2

Let's do part b) since part a) was already answered
by another poster.
If you look in your calculus book, you will find
that the volume of your solid is given by
π ∫ f(x)² dx from 0 to π/3.
This comes from rotating a "typical rectangle" around
the x-axis. The rectangle has length dx and height
f(x). So the "typical cylinder" you get has volume
π f(x)² dx.
Now sum up all these "typical cylinders" to get
the integral.
So we have to find π ∫ tan² x dx from 0 to π/3.
Let's do the indefinite integral and then plug in
all the numbers. We get
∫ tan² x dx = ∫ (sec² x -1) dx = tan x -x + C.
So the volume is given by
π[ tan π/3 - π/3]= π√3 - π²/3.

BTW: Anakin's answer CANNOT be right,
because ln(1/2) is negative! He made a
slip in signs. The correct answer is -ln(1/2) = ln(2).

Answer 3

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