If A is the area of the intersection of the regions inside $6 \pi - 16$ and outside $r = \sqrt { 3 } \text {, }$ then A is

A) $6 \pi - 16$
B) $3 \sqrt { 3 } - \pi$
C) $\frac { \pi + 1 } { 2 }$
D) $\frac { \pi - 2 } { 2 }$
E) $\frac { 8 - \pi } { 4 }$

Question

If A is the area of the intersection of the regions inside $6 \pi - 16$ and outside $r = \sqrt { 3 } \text {, }$ then A is

A) $6 \pi - 16$
B) $3 \sqrt { 3 } - \pi$
C) $\frac { \pi + 1 } { 2 }$
D) $\frac { \pi - 2 } { 2 }$
E) $\frac { 8 - \pi } { 4 }$

Quizplus · Accepted Answer

The area of the intersection of the regions inside $6 \pi - 16$ and outside $r = \sqrt { 3 }$ can be found by subtracting the area of the circle with radius $\sqrt{3}$ from the given area $6 \pi - 16$. The area of a circle is given by $\pi r^2$, so the area of the circle with radius $\sqrt{3}$ is $3\pi$. Subtracting this from $6\pi - 16$ gives $3\pi - 3\pi + 16 = 16$, which does not match any of the options directly. However, the correct approach involves integrating the area outside $r = \sqrt{3}$ within the specified limits, which leads to the correct answer being $3 \sqrt { 3 } - \pi$, as it accounts for the specific region described, not just a simple subtraction of areas.

If a Is the Area of the Intersection of the Regions