Find the area enclosed by the given curves.
-Find the area of the region in the first quadrant bounded on the left by the line $y = \frac { \pi } { 6 }$ and on the right by the curves $y = \tan ^ { 2 } x$ and $y = \cot ^ { 2 } x$. (Round to four decimal places.)
A) 0.5858
B) 0.4126
C) 4.3094
D) 0.3094

Question

Quizplus · Accepted Answer

The area enclosed by the given curves in the first quadrant, bounded on the left by the line $y = \frac { \pi } { 6 }$ and on the right by the curves $y = \tan ^ { 2 } x$ and $y = \cot ^ { 2 } x$, can be found by integrating the difference between the two curves from the intersection point of $y = \tan^2x$ and $y = \cot^2x$ up to the line $y = \frac{\pi}{6}$. The intersection points of $y = \tan^2x$ and $y = \cot^2x$ occur where $\tan^2x = \cot^2x$, which simplifies to $1 = \tan^4x$, or at $x = \frac{\pi}{4}$ in the first quadrant. The area can be calculated by integrating the difference between $y = \cot^2x$ and $y = \tan^2x$ from $x = \frac{\pi}{4}$ to $x = \arctan(\sqrt{\frac{\pi}{6}})$, which gives the result of approximately 0.3094 when rounded to four decimal places.

Find the Area Enclosed by the Given Curves y=π6y = \frac { \pi } { 6 }y=6π​

Find the Area Enclosed by the Given Curves $y = \frac { \pi } { 6 }$