Examples from Change of variables and polar coordinates All examples from Change of variables and polar coordinates.
[ EXAMPLE ] The first-quadrant portion of the unit disk is
0 ≤ r ≤ 1 , 0 ≤ θ ≤ π 2 . 0\le r\le1,\qquad 0\le\theta\le\frac{\pi}{2}. 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π . Its area is
Area ( Ω ) = ∫ 0 π / 2 ∫ 0 1 r d r d θ = ∫ 0 π / 2 1 2 d θ = π 4 . \begin{aligned}
\operatorname{Area}(\Omega)
&=\int_0^{\pi/2}\int_0^1r\,dr\,d\theta\\
&=\int_0^{\pi/2}\frac12\,d\theta
=\frac{\pi}{4}.
\end{aligned} Area ( Ω ) = ∫ 0 π /2 ∫ 0 1 r d r d θ = ∫ 0 π /2 2 1 d θ = 4 π . [RIGHT CLICK] [ASK AI] [ EXAMPLE ] Evaluate
I = ∫ 0 1 ∫ 0 x ( 1 + x 2 + y 2 ) d y d x . I=\int_0^1\int_0^x
\left(1+\sqrt{x^2+y^2}\right)\,dy\,dx. I = ∫ 0 1 ∫ 0 x ( 1 + x 2 + y 2 ) d y d x . The region is the first-quadrant triangle bounded by x = 1 x=1 x = 1 y = x y=x y = x
0 ≤ θ ≤ π 4 , 0 ≤ r ≤ sec θ , 0\le\theta\le\frac{\pi}{4},
\qquad
0\le r\le\sec\theta, 0 ≤ θ ≤ 4 π , 0 ≤ r ≤ sec θ , because x = r cos θ ≤ 1 x=r\cos\theta\le1 x = r cos θ ≤ 1 x 2 + y 2 = r \sqrt{x^2+y^2}=r x 2 + y 2 = r
I = ∫ 0 π / 4 ∫ 0 sec θ ( 1 + r ) r d r d θ = ∫ 0 π / 4 [ r 2 2 + r 3 3 ] 0 sec θ d θ = ∫ 0 π / 4 ( sec 2 θ 2 + sec 3 θ 3 ) d θ = 1 2 + 2 6 + 1 6 ln ( 1 + 2 ) . \begin{aligned}
I
&=\int_0^{\pi/4}\int_0^{\sec\theta}(1+r)r\,dr\,d\theta\\
&=\int_0^{\pi/4}
\left[\frac{r^2}{2}+\frac{r^3}{3}\right]_{0}^{\sec\theta}
d\theta\\
&=\int_0^{\pi/4}
\left(\frac{\sec^2\theta}{2}
+\frac{\sec^3\theta}{3}\right)d\theta\\
&=\frac12+\frac{\sqrt2}{6}
+\frac16\ln(1+\sqrt2).
\end{aligned} I = ∫ 0 π /4 ∫ 0 s e c θ ( 1 + r ) r d r d θ = ∫ 0 π /4 [ 2 r 2 + 3 r 3 ] 0 s e c θ d θ = ∫ 0 π /4 ( 2 sec 2 θ + 3 sec 3 θ ) d θ = 2 1 + 6 2 + 6 1 ln ( 1 + 2 ) . [RIGHT CLICK] [ASK AI] [ EXAMPLE ] The region swept between
r = θ , 0 ≤ θ ≤ 2 π , r=\theta,\qquad0\le\theta\le2\pi, r = θ , 0 ≤ θ ≤ 2 π , and the x x x
0 ≤ θ ≤ 2 π , 0 ≤ r ≤ θ . 0\le\theta\le2\pi,\qquad0\le r\le\theta. 0 ≤ θ ≤ 2 π , 0 ≤ r ≤ θ . Thus
Area ( Ω ) = ∫ 0 2 π ∫ 0 θ r d r d θ = ∫ 0 2 π θ 2 2 d θ = 4 π 3 3 . \begin{aligned}
\operatorname{Area}(\Omega)
&=\int_0^{2\pi}\int_0^\theta r\,dr\,d\theta\\
&=\int_0^{2\pi}\frac{\theta^2}{2}\,d\theta
=\frac{4\pi^3}{3}.
\end{aligned} Area ( Ω ) = ∫ 0 2 π ∫ 0 θ r d r d θ = ∫ 0 2 π 2 θ 2 d θ = 3 4 π 3 . The familiar polar area formula
Area ( Ω ) = 1 2 ∫ r ( θ ) 2 d θ \operatorname{Area}(\Omega)
=\frac12\int r(\theta)^2\,d\theta Area ( Ω ) = 2 1 ∫ r ( θ ) 2 d θ is the result of performing the radial integral first.
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Let Ω \Omega Ω
I = ∬ Ω e − ( x 2 + y 2 ) d A . I=\iint_\Omega e^{-(x^2+y^2)}\,dA. I = ∬ Ω e − ( x 2 + y 2 ) d A . The region is unbounded, so apply the change of variables on bounded
quarter-disks and then let their radius tend to infinity. The integrand is
nonnegative, so monotone convergence justifies this limit. Polar coordinates
give
I = ∫ 0 π / 2 ∫ 0 ∞ e − r 2 r d r d θ = ∫ 0 π / 2 1 2 d θ = π 4 . \begin{aligned}
I
&=\int_0^{\pi/2}\int_0^\infty e^{-r^2}r\,dr\,d\theta\\
&=\int_0^{\pi/2}\frac12\,d\theta
=\frac{\pi}{4}.
\end{aligned} I = ∫ 0 π /2 ∫ 0 ∞ e − r 2 r d r d θ = ∫ 0 π /2 2 1 d θ = 4 π . In Cartesian coordinates the integrand is a product
e − x 2 e − y 2 e^{-x^2}e^{-y^2} e − x 2 e − y 2
I = ( ∫ 0 ∞ e − x 2 d x ) 2 . I
=\left(\int_0^\infty e^{-x^2}\,dx\right)^2. I = ( ∫ 0 ∞ e − x 2 d x ) 2 . Hence
∫ 0 ∞ e − x 2 d x = π 2 . \int_0^\infty e^{-x^2}\,dx=\frac{\sqrt\pi}{2}. ∫ 0 ∞ e − x 2 d x = 2 π . For
G = ∫ − ∞ ∞ e − x 2 / 2 d x , G=\int_{-\infty}^{\infty}e^{-x^2/2}\,dx, G = ∫ − ∞ ∞ e − x 2 /2 d x , symmetry and u = x / 2 u=x/\sqrt2 u = x / 2
G = 2 ∫ 0 ∞ e − x 2 / 2 d x = 2 2 ∫ 0 ∞ e − u 2 d u = 2 π . \begin{aligned}
G
&=2\int_0^\infty e^{-x^2/2}\,dx\\
&=2\sqrt2\int_0^\infty e^{-u^2}\,du
=\sqrt{2\pi}.
\end{aligned} G = 2 ∫ 0 ∞ e − x 2 /2 d x = 2 2 ∫ 0 ∞ e − u 2 d u = 2 π . [RIGHT CLICK] [ASK AI] [ EXAMPLE ] Evaluate
I = ∫ 0 3 ∫ − x x ( 2 − x − y ) 2 d y d x I=\int_0^3\int_{-x}^{x}(2-x-y)^2\,dy\,dx I = ∫ 0 3 ∫ − x x ( 2 − x − y ) 2 d y d x using
u = x + y , v = x − y . u=x+y,\qquad v=x-y. u = x + y , v = x − y . The inverse transformation is
x = u + v 2 , y = u − v 2 , x=\frac{u+v}{2},
\qquad
y=\frac{u-v}{2}, x = 2 u + v , y = 2 u − v , and
∣ ∂ ( x , y ) ∂ ( u , v ) ∣ = ∣ det ( 1 2 1 2 1 2 − 1 2 ) ∣ = 1 2 . \left|\frac{\partial(x,y)}{\partial(u,v)}\right|
=
\left|
\det
\begin{pmatrix}
\frac12&\frac12\\[2pt]
\frac12&-\frac12
\end{pmatrix}
\right|
=\frac12. ∂ ( u , v ) ∂ ( x , y ) = det ( 2 1 2 1 2 1 − 2 1 ) = 2 1 . The original region is bounded by y = − x y=-x y = − x y = x y=x y = x x = 3 x=3 x = 3
y = − x ⟹ u = 0 , y = x ⟹ v = 0 , x = 3 ⟹ u + v = 6. y=-x\Longrightarrow u=0,\qquad
y=x\Longrightarrow v=0,\qquad
x=3\Longrightarrow u+v=6. y = − x ⟹ u = 0 , y = x ⟹ v = 0 , x = 3 ⟹ u + v = 6. Thus
0 ≤ u ≤ 6 , 0 ≤ v ≤ 6 − u . 0\le u\le6,\qquad0\le v\le6-u. 0 ≤ u ≤ 6 , 0 ≤ v ≤ 6 − u . Since x + y = u x+y=u x + y = u ( 2 − u ) 2 (2-u)^2 ( 2 − u ) 2
I = ∫ 0 6 ∫ 0 6 − u ( 2 − u ) 2 1 2 d v d u = 1 2 ∫ 0 6 ( 2 − u ) 2 ( 6 − u ) d u = 18. \begin{aligned}
I
&=\int_0^6\int_0^{6-u}(2-u)^2\frac12\,dv\,du\\
&=\frac12\int_0^6(2-u)^2(6-u)\,du
=18.
\end{aligned} I = ∫ 0 6 ∫ 0 6 − u ( 2 − u ) 2 2 1 d v d u = 2 1 ∫ 0 6 ( 2 − u ) 2 ( 6 − u ) d u = 18. These coordinates suit the problem because x + y x+y x + y
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Evaluate
∬ Ω ( 4 x 2 − 2 y 2 ) d A , \iint_\Omega(4x^2-2y^2)\,dA, ∬ Ω ( 4 x 2 − 2 y 2 ) d A , where
Ω = { ( x , y ) : 1 ≤ x y ≤ 4 , x ≤ y ≤ 4 x } . \Omega=\{(x,y):1\le xy\le4,\ x\le y\le4x\}. Ω = {( x , y ) : 1 ≤ x y ≤ 4 , x ≤ y ≤ 4 x } . Use
u = x y , v = y x . u=xy,\qquad v=\frac yx. u = x y , v = x y . On Ω \Omega Ω x > 0 x>0 x > 0 x < 0 x<0 x < 0 4 x < x 4x<x 4 x < x x ≤ y ≤ 4 x x\le y\le4x x ≤ y ≤ 4 x
x = u v , y = u v , x=\sqrt{\frac uv},\qquad y=\sqrt{uv}, x = v u , y = uv , for u , v > 0 u,v>0 u , v > 0
1 ≤ u ≤ 4 , 1 ≤ v ≤ 4. 1\le u\le4,\qquad1\le v\le4. 1 ≤ u ≤ 4 , 1 ≤ v ≤ 4. The forward Jacobian is
∂ ( u , v ) ∂ ( x , y ) = det ( y x − y / x 2 1 / x ) = y x + y x = 2 v . \frac{\partial(u,v)}{\partial(x,y)}
=\det
\begin{pmatrix}
y&x\\[2pt]
-y/x^2&1/x
\end{pmatrix}
=\frac yx+\frac yx=2v. ∂ ( x , y ) ∂ ( u , v ) = det ( y − y / x 2 x 1/ x ) = x y + x y = 2 v . Hence
d A = ∣ ∂ ( x , y ) ∂ ( u , v ) ∣ d u d v = 1 2 v d u d v . dA
=\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv
=\frac1{2v}\,du\,dv. d A = ∂ ( u , v ) ∂ ( x , y ) d u d v = 2 v 1 d u d v . Since x 2 = u / v x^2=u/v x 2 = u / v y 2 = u v y^2=uv y 2 = uv
4 x 2 − 2 y 2 = 4 u v − 2 u v . 4x^2-2y^2=\frac{4u}{v}-2uv. 4 x 2 − 2 y 2 = v 4 u − 2 uv . Therefore
I = ∫ 1 4 ∫ 1 4 ( 4 u v − 2 u v ) 1 2 v d v d u = ∫ 1 4 ∫ 1 4 ( 2 u v 2 − u ) d v d u = ∫ 1 4 [ − 2 u v − u v ] 1 4 d u = ∫ 1 4 − 3 u 2 d u = − 45 4 . \begin{aligned}
I
&=\int_1^4\int_1^4
\left(\frac{4u}{v}-2uv\right)\frac1{2v}\,dv\,du\\
&=\int_1^4\int_1^4\left(\frac{2u}{v^2}-u\right)\,dv\,du\\
&=\int_1^4\left[-\frac{2u}{v}-uv\right]_{1}^{4}\,du\\
&=\int_1^4-\frac{3u}{2}\,du
=-\frac{45}{4}.
\end{aligned} I = ∫ 1 4 ∫ 1 4 ( v 4 u − 2 uv ) 2 v 1 d v d u = ∫ 1 4 ∫ 1 4 ( v 2 2 u − u ) d v d u = ∫ 1 4 [ − v 2 u − uv ] 1 4 d u = ∫ 1 4 − 2 3 u d u = − 4 45 . The answer is negative because the integrand takes both signs; a double
integral is not automatically a geometric volume.
[RIGHT CLICK] [ASK AI] Change of variables and polar coordinates Leads to Seen Questions and 2 more.