Answer to Question #111844 in Calculus for sohaib
2020-04-23T17:49:01-04:00
Use a double integral to derive the area of the region between circles of radius a and b with
α
≤
θ
≤
β
α≤θ≤β
. See the image below for a sketch of the region.
1
2020-04-29T16:37:30-0400
S = ∬ a 2 ≤ x 2 + y 2 ≤ b 2 α ≤ θ ≤ β 1 d x d y = [ x = r ⋅ cos θ y = r ⋅ sin θ d x d y = r d r d θ r ∈ [ a ; b ] α ≤ θ ≤ β ] = = ∫ α β d θ ⋅ ( ∫ a b r d r ) = θ ∣ α β ⋅ ( r 2 2 ∣ a b ) = ( β − α ) ⋅ b 2 − a 2 2 S=\iint\limits_{\begin{matrix}
a^2\le x^2+y^2\le b^2\\
\alpha\le\theta\le\beta
\end{matrix}}1dxdy
=\left[\begin{array}{l}
x=r\cdot\cos\theta\\
y=r\cdot\sin\theta\\
dxdy=rdrd\theta\\
r\in[a;b]\\
\alpha\le\theta\le\beta
\end{array}\right]=\\[0.3cm]
=\int\limits_\alpha^\beta d\theta\cdot\left(\int\limits_a^brdr\right)=\left.\theta\right|_\alpha^\beta\cdot\left(\left.\frac{r^2}{2}\right|_a^b\right)=\left(\beta-\alpha\right)\cdot\frac{b^2-a^2}{2} S = a 2 ≤ x 2 + y 2 ≤ b 2 α ≤ θ ≤ β ∬ 1 d x d y = ⎣ ⎡ x = r ⋅ cos θ y = r ⋅ sin θ d x d y = r d r d θ r ∈ [ a ; b ] α ≤ θ ≤ β ⎦ ⎤ = = α ∫ β d θ ⋅ ( a ∫ b r d r ) = θ ∣ α β ⋅ ( 2 r 2 ∣ ∣ a b ) = ( β − α ) ⋅ 2 b 2 − a 2
Conclusion,
S = ( β − α ) ⋅ ( b 2 − a 2 ) 2 \boxed{S=\frac{\left(\beta-\alpha\right)\cdot\left(b^2-a^2\right)}{2}} S = 2 ( β − α ) ⋅ ( b 2 − a 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
Leave a comment