← 2015 Paper 1
UPSC 2015 Maths Optional Paper 1 Q4a — Step-by-Step Solution 13 marks · Section A
Double integrals · Calculus · asked 10× in 13 yrs · Read the full method →
Question
Evaluate ∬ R ∣ y − x 2 ∣ d x d y \displaystyle\iint_R \sqrt{|y-x^2|}\,dx\,dy ∬ R ∣ y − x 2 ∣ d x d y R = [ − 1 , 1 ] × [ 0 , 2 ] R=[-1,1]\times[0,2] R = [ − 1 , 1 ] × [ 0 , 2 ]
Technique
Split the rectangle along y = x 2 y=x^2 y = x 2 y y y 2 3 ( half-width ) 3 / 2 \dfrac{2}{3}(\text{half-width})^{3/2} 3 2 ( half-width ) 3/2 x x x ( 2 − x 2 ) 3 / 2 (2-x^2)^{3/2} ( 2 − x 2 ) 3/2 x = 2 sin θ x=\sqrt 2\sin\theta x = 2 sin θ cos 4 \cos^4 cos 4
Solution
Strategy. The parabola y = x 2 y=x^2 y = x 2 R R R y < x 2 y<x^2 y < x 2 ∣ y − x 2 ∣ = x 2 − y |y-x^2|=x^2-y ∣ y − x 2 ∣ = x 2 − y y ≥ x 2 y\ge x^2 y ≥ x 2 ∣ y − x 2 ∣ = y − x 2 |y-x^2|=y-x^2 ∣ y − x 2 ∣ = y − x 2 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] 0 ≤ x 2 ≤ 1 ≤ 2 0\le x^2\le 1\le 2 0 ≤ x 2 ≤ 1 ≤ 2 R R R x ∈ [ − 1 , 1 ] x\in[-1,1] x ∈ [ − 1 , 1 ] y y y 0 0 0 x 2 x^2 x 2 x 2 x^2 x 2 2 2 2
Also the integrand is even in x x x x 2 x^2 x 2 y y y x x x [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] 2 × 2\times 2 × x x x [ 0 , 1 ] [0,1] [ 0 , 1 ]
Step 1 — Split
I = ∫ − 1 1 ∫ 0 2 ∣ y − x 2 ∣ d y d x = ∫ − 1 1 [ I 1 ( x ) + I 2 ( x ) ] d x , I=\int_{-1}^{1}\!\int_0^2\sqrt{|y-x^2|}\,dy\,dx=\int_{-1}^{1}\bigl[I_1(x)+I_2(x)\bigr]\,dx, I = ∫ − 1 1 ∫ 0 2 ∣ y − x 2 ∣ d y d x = ∫ − 1 1 [ I 1 ( x ) + I 2 ( x ) ] d x , where
I 1 ( x ) = ∫ 0 x 2 x 2 − y d y I_1(x)=\int_0^{x^2}\sqrt{x^2-y}\,dy I 1 ( x ) = ∫ 0 x 2 x 2 − y d y I 2 ( x ) = ∫ x 2 2 y − x 2 d y I_2(x)=\int_{x^2}^{2}\sqrt{y-x^2}\,dy I 2 ( x ) = ∫ x 2 2 y − x 2 d y
Step 2 — Compute I 1 ( x ) I_1(x) I 1 ( x )
Substitute u = x 2 − y u=x^2-y u = x 2 − y d u = − d y du=-dy d u = − d y y = 0 y=0 y = 0 u = x 2 u=x^2 u = x 2 y = x 2 y=x^2 y = x 2 u = 0 u=0 u = 0
I 1 ( x ) = ∫ x 2 0 u ( − d u ) = ∫ 0 x 2 u d u = 2 3 ( x 2 ) 3 / 2 = 2 3 ∣ x ∣ 3 . I_1(x)=\int_{x^2}^{0}\sqrt{u}(-du)=\int_0^{x^2}\sqrt{u}\,du=\dfrac{2}{3}(x^2)^{3/2}=\dfrac{2}{3}|x|^3. I 1 ( x ) = ∫ x 2 0 u ( − d u ) = ∫ 0 x 2 u d u = 3 2 ( x 2 ) 3/2 = 3 2 ∣ x ∣ 3 . Step 3 — Compute I 2 ( x ) I_2(x) I 2 ( x )
Substitute u = y − x 2 u=y-x^2 u = y − x 2 d u = d y du=dy d u = d y y = x 2 y=x^2 y = x 2 u = 0 u=0 u = 0 y = 2 y=2 y = 2 u = 2 − x 2 u=2-x^2 u = 2 − x 2
I 2 ( x ) = ∫ 0 2 − x 2 u d u = 2 3 ( 2 − x 2 ) 3 / 2 . I_2(x)=\int_0^{2-x^2}\sqrt{u}\,du=\dfrac{2}{3}(2-x^2)^{3/2}. I 2 ( x ) = ∫ 0 2 − x 2 u d u = 3 2 ( 2 − x 2 ) 3/2 . Step 4 — Combine and integrate over x x x
I = ∫ − 1 1 [ 2 3 ∣ x ∣ 3 + 2 3 ( 2 − x 2 ) 3 / 2 ] d x = 2 3 ∫ − 1 1 ∣ x ∣ 3 d x + 2 3 ∫ − 1 1 ( 2 − x 2 ) 3 / 2 d x . I=\int_{-1}^{1}\left[\dfrac{2}{3}|x|^3+\dfrac{2}{3}(2-x^2)^{3/2}\right]\,dx=\dfrac{2}{3}\int_{-1}^{1}|x|^3\,dx+\dfrac{2}{3}\int_{-1}^{1}(2-x^2)^{3/2}\,dx. I = ∫ − 1 1 [ 3 2 ∣ x ∣ 3 + 3 2 ( 2 − x 2 ) 3/2 ] d x = 3 2 ∫ − 1 1 ∣ x ∣ 3 d x + 3 2 ∫ − 1 1 ( 2 − x 2 ) 3/2 d x . First piece. ∫ − 1 1 ∣ x ∣ 3 d x = 2 ∫ 0 1 x 3 d x = 2 ⋅ 1 4 = 1 2 \int_{-1}^{1}|x|^3\,dx=2\int_0^1 x^3\,dx=2\cdot\dfrac{1}{4}=\dfrac{1}{2} ∫ − 1 1 ∣ x ∣ 3 d x = 2 ∫ 0 1 x 3 d x = 2 ⋅ 4 1 = 2 1 2 3 ⋅ 1 2 = 1 3 \dfrac{2}{3}\cdot\dfrac{1}{2}=\dfrac{1}{3} 3 2 ⋅ 2 1 = 3 1
Second piece. J = ∫ − 1 1 ( 2 − x 2 ) 3 / 2 d x = 2 ∫ 0 1 ( 2 − x 2 ) 3 / 2 d x J=\int_{-1}^{1}(2-x^2)^{3/2}\,dx=2\int_0^1(2-x^2)^{3/2}\,dx J = ∫ − 1 1 ( 2 − x 2 ) 3/2 d x = 2 ∫ 0 1 ( 2 − x 2 ) 3/2 d x
Substitute x = 2 sin θ x=\sqrt{2}\sin\theta x = 2 sin θ d x = 2 cos θ d θ dx=\sqrt 2\cos\theta\,d\theta d x = 2 cos θ d θ x = 0 x=0 x = 0 θ = 0 \theta=0 θ = 0 x = 1 x=1 x = 1 θ = π / 4 \theta=\pi/4 θ = π /4
2 − x 2 = 2 cos 2 θ 2-x^2=2\cos^2\theta 2 − x 2 = 2 cos 2 θ ( 2 − x 2 ) 3 / 2 = ( 2 cos 2 θ ) 3 / 2 = 2 3 / 2 cos 3 θ = 2 2 cos 3 θ (2-x^2)^{3/2}=(2\cos^2\theta)^{3/2}=2^{3/2}\cos^3\theta=2\sqrt 2\cos^3\theta ( 2 − x 2 ) 3/2 = ( 2 cos 2 θ ) 3/2 = 2 3/2 cos 3 θ = 2 2 cos 3 θ
∫ 0 1 ( 2 − x 2 ) 3 / 2 d x = ∫ 0 π / 4 2 2 cos 3 θ ⋅ 2 cos θ d θ = 4 ∫ 0 π / 4 cos 4 θ d θ . \int_0^1(2-x^2)^{3/2}\,dx=\int_0^{\pi/4}2\sqrt 2\cos^3\theta\cdot\sqrt 2\cos\theta\,d\theta=4\int_0^{\pi/4}\cos^4\theta\,d\theta. ∫ 0 1 ( 2 − x 2 ) 3/2 d x = ∫ 0 π /4 2 2 cos 3 θ ⋅ 2 cos θ d θ = 4 ∫ 0 π /4 cos 4 θ d θ . Use cos 4 θ = 3 + 4 cos 2 θ + cos 4 θ 8 \cos^4\theta=\dfrac{3+4\cos 2\theta+\cos 4\theta}{8} cos 4 θ = 8 3 + 4 cos 2 θ + cos 4 θ
∫ 0 π / 4 cos 4 θ d θ = 1 8 ∫ 0 π / 4 [ 3 + 4 cos 2 θ + cos 4 θ ] d θ \int_0^{\pi/4}\cos^4\theta\,d\theta=\dfrac{1}{8}\int_0^{\pi/4}\bigl[3+4\cos 2\theta+\cos 4\theta\bigr]\,d\theta ∫ 0 π /4 cos 4 θ d θ = 8 1 ∫ 0 π /4 [ 3 + 4 cos 2 θ + cos 4 θ ] d θ = 1 8 [ 3 θ + 2 sin 2 θ + sin 4 θ 4 ] 0 π / 4 =\dfrac{1}{8}\left[3\theta+2\sin 2\theta+\dfrac{\sin 4\theta}{4}\right]_0^{\pi/4} = 8 1 [ 3 θ + 2 sin 2 θ + 4 sin 4 θ ] 0 π /4 = 1 8 [ 3 π 4 + 2 sin ( π / 2 ) + sin π 4 ] = 1 8 [ 3 π 4 + 2 + 0 ] = 3 π 32 + 1 4 . =\dfrac{1}{8}\left[\dfrac{3\pi}{4}+2\sin(\pi/2)+\dfrac{\sin\pi}{4}\right]=\dfrac{1}{8}\left[\dfrac{3\pi}{4}+2+0\right]=\dfrac{3\pi}{32}+\dfrac{1}{4}. = 8 1 [ 4 3 π + 2 sin ( π /2 ) + 4 sin π ] = 8 1 [ 4 3 π + 2 + 0 ] = 32 3 π + 4 1 . Thus ∫ 0 1 ( 2 − x 2 ) 3 / 2 d x = 4 ( 3 π 32 + 1 4 ) = 3 π 8 + 1 \int_0^1(2-x^2)^{3/2}dx=4\bigl(\dfrac{3\pi}{32}+\dfrac{1}{4}\bigr)=\dfrac{3\pi}{8}+1 ∫ 0 1 ( 2 − x 2 ) 3/2 d x = 4 ( 32 3 π + 4 1 ) = 8 3 π + 1
So J = 2 ( 3 π 8 + 1 ) = 3 π 4 + 2 J=2\bigl(\dfrac{3\pi}{8}+1\bigr)=\dfrac{3\pi}{4}+2 J = 2 ( 8 3 π + 1 ) = 4 3 π + 2
Second piece total: 2 3 J = 2 3 ( 3 π 4 + 2 ) = π 2 + 4 3 \dfrac{2}{3}J=\dfrac{2}{3}\bigl(\dfrac{3\pi}{4}+2\bigr)=\dfrac{\pi}{2}+\dfrac{4}{3} 3 2 J = 3 2 ( 4 3 π + 2 ) = 2 π + 3 4
Step 5 — Add
I = 1 3 + π 2 + 4 3 = 5 3 + π 2 . I=\dfrac{1}{3}+\dfrac{\pi}{2}+\dfrac{4}{3}=\dfrac{5}{3}+\dfrac{\pi}{2}. I = 3 1 + 2 π + 3 4 = 3 5 + 2 π . Answer
I = 5 3 + π 2 . \boxed{\;I=\dfrac{5}{3}+\dfrac{\pi}{2}.\;} I = 3 5 + 2 π .