Find the area of the part of the cylinder $y ^ { 2 } + z ^ { 2 } = 9$ that is above the rectangle R = [0, 2] $\times$ [3, 3].
A)3$\pi$
B)2$\pi$
C)3
D)2
E)12$\pi$
F)6$\pi$
G)12
H)6

Question

Quizplus · Accepted Answer

The area of the cylinder above the rectangle is calculated by integrating the length of the cylinder's intersection with the rectangle along the x-axis. The length of the intersection is the circumference of the circle $y^2 + z^2 = 9$, which is $6\pi$. The x-interval is [0, 2], so the area is $6\pi \times 2 = 12\pi$.

Find the Area of the Part of the Cylinder y2+z2=9y ^ { 2 } + z ^ { 2 } = 9y2+z2=9

Find the Area of the Part of the Cylinder $y ^ { 2 } + z ^ { 2 } = 9$