Answer
$50\pi$
Work Step by Step
$x=5cosθ$, $y=5sinθ$, $0\leq θ \leq \frac{\pi}{2}$, y-axis
Use the area for the surface area of a parametric curve:
$S=2\pi\int_{a}^{b}x(θ)\sqrt ((\frac{dx}{dθ})^2+(\frac{dy}{dθ})^2)dθ$
$\frac{dx}{dθ}=-5sinθ$
$\frac{dy}{dθ}=5cosθ$
$S=2\pi\int_{0}^{\frac{\pi}{2}}5cosθ\sqrt ((-5sinθ)^2+(5cosθ)^2)dθ$
$S=2\pi\int_{0}^{\frac{\pi}{2}}5cosθ\sqrt (25sin^2θ+25cos^2θ)dθ$
$S=2\pi\int_{0}^{\frac{\pi}{2}}5cosθ\sqrt (25(sin^2θ+cos^2θ)dθ$
By the Pythagorean identity $sin^2θ+cos^2θ=1$:
$S=2\pi\int_{0}^{\frac{\pi}{2}}25cosθdθ$
$S=50\pi\int_{0}^{\frac{\pi}{2}}cosθdθ$
$S=50\pi[sinθ]_{0}^{\frac{\pi}{2}}$
$S=50\pi[1-0]$
$S=50\pi$
