Question #346263

find the surface area of the portion of the curve x^2+y^2=4 from x=0 to x=2 when it is revolved about the y-axis


1
Expert's answer
2022-05-30T23:07:53-0400
S=2πabrcos(t)(r(t))2+(r(t))2dtS=2\pi\displaystyle\int_{a}^{b}r\cos (t)\sqrt{(r(t))^2+(r'(t))^2}dt

x2+y2=4x^2+y^2=4

r=2,a=π/2,b=π/2r=2, a=-\pi/2, b=\pi/2

S=2ππ/2π/22cost(2)2+(0)2dtS=2\pi\displaystyle\int_{-\pi/2}^{\pi/2}2\cos t\sqrt{(2)^2+(0)^2}dt

=8π[sint]π/2π/2=16π(units2)=8\pi[\sin t]\begin{matrix} \pi/2 \\ -\pi/2 \end{matrix}=16\pi({units}^2)


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS