What is the area enclosed by r=sintheta/theta-theta^3-theta between theta in [pi/12,pi/3]?

1 Answer
May 26, 2017

Signed area ~~-0.08885...

Explanation:

Use the Riemann integral.
Area under f(x) from [a, b] is int_a^bf(x) dx

So, the area under r(theta)=(sintheta)/theta-theta^3-theta where theta in [pi/12, pi/3] is int_(pi/12)^(pi/3)r(theta) \quad d theta

Let's integrate the indefinite and ignore the constant:

int \quad r(theta) \quad d theta
=int\quad (sintheta)/theta-theta^3-theta \quad d theta
=int\quad (sintheta)/theta \quad d theta - int\quad theta^3 \quad d theta - int\quad theta \quad d theta

Now, int\quadsintheta/theta\quadd theta="Si"(theta)+C which is a very interesting function. We have entered the lair of Trigonometric Integrals. There is no closed form of "Si", so we would have to deal with "Si"(pi/3)-"Si"(pi/12)~~0.724654...

Anyways, we now integrate the (trivial) rest, again, ignoring the constant

int\quad theta^3 \quad d theta=(theta^4)/4

int\quad theta \quad d theta=(theta^2)/2

Finally, we piece everything together:
int_(pi/12)^(pi/3)r(theta) \quad d theta=|["Si"(theta)-((theta^2)(theta^2+2))/4]|_(pi/12)^(pi/3)
~~0.724564...-((pi/3)^2((pi/3)^2+2))/4+((pi/12)^2((pi/12)^2+2))/4
~~0.724564...-0.813512...

~~-0.08885...

:. The (unsigned) area under r(theta) is approx. 0.0889