calculate ∫∫_((x^2 /a^2 ) +(y^2 /b^2 ) ≤1) (x^2 −y^2 )dxdy whit a>0 and b>0 .


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 43538 by abdo.msup.com last updated on 11/Sep/18

$c a l c u l a t e \int \int_{\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} ⩽ 1} (x^{2} - y^{2}) d x d y w h i t$ $a > 0 a n d b > 0 .$
Commented bymaxmathsup by imad last updated on 16/Sep/18
$let consider the diffeomorphism (r,θ) →(x,y)=(arcosθ,brsinθ) I = ∫∫_(0≤r≤1 and 0≤θ≤2π) (a^2 cos^2 θ −b^2 sin^2 θ)ab r dr dθ =a^3 b∫_0 ^1 rdr ∫_0 ^(2π) cos^2 θ dθ −ab^3 ∫_0 ^1 rdr ∫_0 ^(2π) sin^2 θ dθ =((a^3 b)/2) ∫_0 ^(2π) ((1+cos(2θ))/2)dθ −((ab^3 )/2) ∫_0 ^(2π) ((1−cos(2θ))/2) dθ A=((π a^3 b)/2) +((a^3 b)/4) ∫_0 ^(2π) cos(2θ)dθ −((πab^3 )/2) + ((ab^3 )/4) ∫_0 ^(2π) cos(2θ)dθ =((πab)/2)(a^2 −b^2 ) +0+0 ⇒A =((πab)/2){a^2 −b^2 } .$
$l e t c o n s i d e r t h e d i f f e o m o r p h i s m (r, θ) \to (x, y) = (a r c o s θ, b r s i n θ)$ $I = \int \int_{0 ⩽ r ⩽ 1 a n d 0 ⩽ θ ⩽ 2 π} (a^{2} {c o s}^{2} θ - b^{2} {s i n}^{2} θ) a b r d r d θ$ $= a^{3} b \int_{0}^{1} r d r \int_{0}^{2 π} {c o s}^{2} θ d θ - {a b}^{3} \int_{0}^{1} r d r \int_{0}^{2 π} {s i n}^{2} θ d θ$ $= \frac{a^{3} b}{2} \int_{0}^{2 π} \frac{1 + c o s (2 θ)}{2} d θ - \frac{{a b}^{3}}{2} \int_{0}^{2 π} \frac{1 - c o s (2 θ)}{2} d θ$ $A = \frac{π a^{3} b}{2} + \frac{a^{3} b}{4} \int_{0}^{2 π} c o s (2 θ) d θ - \frac{π {a b}^{3}}{2} + \frac{{a b}^{3}}{4} \int_{0}^{2 π} c o s (2 θ) d θ$ $= \frac{π a b}{2} (a^{2} - b^{2}) + 0 + 0 \Rightarrow A = \frac{π a b}{2} {a^{2} - b^{2}} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum