PYQs / GATE EC / 2026 / Set 1 / Q61 GATE EC 2026 Set 1 — Question 61 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +2 / -0 Medium Double & Triple Integrals Calculus Engineering Mathematics
Engineering Mathematics → Calculus → Double & Triple Integrals
Last updated 5 September 2026
Question Consider the square region
R R R in the
X − Y X-Y X − Y plane as shown with the dark shading in the Figure. The value of
∬ R ( x 2 + y 2 − 1 ) d x d y \iint_R (x^2 + y^2 - 1) dx dy ∬ R ( x 2 + y 2 − 1 ) d x d y is
____ .
(rounded off to two decimal places)
Solution The shaded region
R R R is a square with vertices at
( 1 , 0 ) , ( 2 , 1 ) , ( 1 , 2 ) , ( 0 , 1 ) (1,0), (2,1), (1,2), (0,1) ( 1 , 0 ) , ( 2 , 1 ) , ( 1 , 2 ) , ( 0 , 1 ) .
To evaluate the integral
I = ∬ R ( x 2 + y 2 − 1 ) d x d y I = \iint_R (x^2 + y^2 - 1) dx dy I = ∬ R ( x 2 + y 2 − 1 ) d x d y , we use the transformation:
u = x + y u = x + y u = x + y and
v = x − y v = x - y v = x − y .
The vertices in the
( u , v ) (u,v) ( u , v ) plane are:
( 1 , 0 ) → ( 1 , 1 ) (1,0) \rightarrow (1,1) ( 1 , 0 ) → ( 1 , 1 ) ( 2 , 1 ) → ( 3 , 1 ) (2,1) \rightarrow (3,1) ( 2 , 1 ) → ( 3 , 1 ) ( 1 , 2 ) → ( 3 , − 1 ) (1,2) \rightarrow (3,-1) ( 1 , 2 ) → ( 3 , − 1 ) ( 0 , 1 ) → ( 1 , − 1 ) (0,1) \rightarrow (1,-1) ( 0 , 1 ) → ( 1 , − 1 ) Thus, the limits are
1 ≤ u ≤ 3 1 \leq u \leq 3 1 ≤ u ≤ 3 and
− 1 ≤ v ≤ 1 -1 \leq v \leq 1 − 1 ≤ v ≤ 1 .
The Jacobian of the transformation is
∣ J ∣ = ∣ ∂ ( x , y ) ∂ ( u , v ) ∣ = 1 2 |J| = \left| \frac{\partial(x,y)}{\partial(u,v)} \right| = \frac{1}{2} ∣ J ∣ = ∂ ( u , v ) ∂ ( x , y ) = 2 1 .
Also,
x 2 + y 2 = ( u + v ) 2 4 + ( u − v ) 2 4 = u 2 + v 2 2 x^2 + y^2 = \frac{(u+v)^2}{4} + \frac{(u-v)^2}{4} = \frac{u^2 + v^2}{2} x 2 + y 2 = 4 ( u + v ) 2 + 4 ( u − v ) 2 = 2 u 2 + v 2 .
The integral becomes:
I = ∫ − 1 1 ∫ 1 3 ( u 2 + v 2 2 − 1 ) 1 2 d u d v = 1 4 ∫ − 1 1 [ u 3 3 + u v 2 − 2 u ] 1 3 d v I = \int_{-1}^{1} \int_{1}^{3} \left( \frac{u^2 + v^2}{2} - 1 \right) \frac{1}{2} du dv = \frac{1}{4} \int_{-1}^{1} \left[ \frac{u^3}{3} + uv^2 - 2u \right]_{1}^{3} dv I = ∫ − 1 1 ∫ 1 3 ( 2 u 2 + v 2 − 1 ) 2 1 d u d v = 4 1 ∫ − 1 1 [ 3 u 3 + u v 2 − 2 u ] 1 3 d v I = 1 4 ∫ − 1 1 ( ( 9 + 3 v 2 − 6 ) − ( 1 3 + v 2 − 2 ) ) d v = 1 4 ∫ − 1 1 ( 2 v 2 + 14 3 ) d v I = \frac{1}{4} \int_{-1}^{1} \left( (9 + 3v^2 - 6) - (\frac{1}{3} + v^2 - 2) \right) dv = \frac{1}{4} \int_{-1}^{1} (2v^2 + \frac{14}{3}) dv I = 4 1 ∫ − 1 1 ( ( 9 + 3 v 2 − 6 ) − ( 3 1 + v 2 − 2 ) ) d v = 4 1 ∫ − 1 1 ( 2 v 2 + 3 14 ) d v I = 1 4 [ 2 v 3 3 + 14 v 3 ] − 1 1 = 1 4 ( ( 2 3 + 14 3 ) − ( − 2 3 − 14 3 ) ) = 1 4 ( 16 3 + 16 3 ) = 8 3 ≈ 2.67 I = \frac{1}{4} \left[ \frac{2v^3}{3} + \frac{14v}{3} \right]_{-1}^{1} = \frac{1}{4} \left( (\frac{2}{3} + \frac{14}{3}) - (-\frac{2}{3} - \frac{14}{3}) \right) = \frac{1}{4} \left( \frac{16}{3} + \frac{16}{3} \right) = \frac{8}{3} \approx 2.67 I = 4 1 [ 3 2 v 3 + 3 14 v ] − 1 1 = 4 1 ( ( 3 2 + 3 14 ) − ( − 3 2 − 3 14 ) ) = 4 1 ( 3 16 + 3 16 ) = 3 8 ≈ 2.67 .
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Solution The shaded region
R R R is a square with vertices at
( 1 , 0 ) , ( 2 , 1 ) , ( 1 , 2 ) , ( 0 , 1 ) (1,0), (2,1), (1,2), (0,1) ( 1 , 0 ) , ( 2 , 1 ) , ( 1 , 2 ) , ( 0 , 1 ) .
To evaluate the integral
I = ∬ R ( x 2 + y 2 − 1 ) d x d y I = \iint_R (x^2 + y^2 - 1) dx dy I = ∬ R ( x 2 + y 2 − 1 ) d x d y , we use the transformation:
u = x + y u = x + y u = x + y and
v = x − y v = x - y v = x − y .
The vertices in the
( u , v ) (u,v) ( u , v ) plane are:
( 1 , 0 ) → ( 1 , 1 ) (1,0) \rightarrow (1,1) ( 1 , 0 ) → ( 1 , 1 ) ( 2 , 1 ) → ( 3 , 1 ) (2,1) \rightarrow (3,1) ( 2 , 1 ) → ( 3 , 1 ) ( 1 , 2 ) → ( 3 , − 1 ) (1,2) \rightarrow (3,-1) ( 1 , 2 ) → ( 3 , − 1 ) ( 0 , 1 ) → ( 1 , − 1 ) (0,1) \rightarrow (1,-1) ( 0 , 1 ) → ( 1 , − 1 ) Thus, the limits are
1 ≤ u ≤ 3 1 \leq u \leq 3 1 ≤ u ≤ 3 and
− 1 ≤ v ≤ 1 -1 \leq v \leq 1 − 1 ≤ v ≤ 1 .
The Jacobian of the transformation is
∣ J ∣ = ∣ ∂ ( x , y ) ∂ ( u , v ) ∣ = 1 2 |J| = \left| \frac{\partial(x,y)}{\partial(u,v)} \right| = \frac{1}{2} ∣ J ∣ = ∂ ( u , v ) ∂ ( x , y ) = 2 1 .
Also,
x 2 + y 2 = ( u + v ) 2 4 + ( u − v ) 2 4 = u 2 + v 2 2 x^2 + y^2 = \frac{(u+v)^2}{4} + \frac{(u-v)^2}{4} = \frac{u^2 + v^2}{2} x 2 + y 2 = 4 ( u + v ) 2 + 4 ( u − v ) 2 = 2 u 2 + v 2 .
The integral becomes:
I = ∫ − 1 1 ∫ 1 3 ( u 2 + v 2 2 − 1 ) 1 2 d u d v = 1 4 ∫ − 1 1 [ u 3 3 + u v 2 − 2 u ] 1 3 d v I = \int_{-1}^{1} \int_{1}^{3} \left( \frac{u^2 + v^2}{2} - 1 \right) \frac{1}{2} du dv = \frac{1}{4} \int_{-1}^{1} \left[ \frac{u^3}{3} + uv^2 - 2u \right]_{1}^{3} dv I = ∫ − 1 1 ∫ 1 3 ( 2 u 2 + v 2 − 1 ) 2 1 d u d v = 4 1 ∫ − 1 1 [ 3 u 3 + u v 2 − 2 u ] 1 3 d v I = 1 4 ∫ − 1 1 ( ( 9 + 3 v 2 − 6 ) − ( 1 3 + v 2 − 2 ) ) d v = 1 4 ∫ − 1 1 ( 2 v 2 + 14 3 ) d v I = \frac{1}{4} \int_{-1}^{1} \left( (9 + 3v^2 - 6) - (\frac{1}{3} + v^2 - 2) \right) dv = \frac{1}{4} \int_{-1}^{1} (2v^2 + \frac{14}{3}) dv I = 4 1 ∫ − 1 1 ( ( 9 + 3 v 2 − 6 ) − ( 3 1 + v 2 − 2 ) ) d v = 4 1 ∫ − 1 1 ( 2 v 2 + 3 14 ) d v I = 1 4 [ 2 v 3 3 + 14 v 3 ] − 1 1 = 1 4 ( ( 2 3 + 14 3 ) − ( − 2 3 − 14 3 ) ) = 1 4 ( 16 3 + 16 3 ) = 8 3 ≈ 2.67 I = \frac{1}{4} \left[ \frac{2v^3}{3} + \frac{14v}{3} \right]_{-1}^{1} = \frac{1}{4} \left( (\frac{2}{3} + \frac{14}{3}) - (-\frac{2}{3} - \frac{14}{3}) \right) = \frac{1}{4} \left( \frac{16}{3} + \frac{16}{3} \right) = \frac{8}{3} \approx 2.67 I = 4 1 [ 3 2 v 3 + 3 14 v ] − 1 1 = 4 1 ( ( 3 2 + 3 14 ) − ( − 3 2 − 3 14 ) ) = 4 1 ( 3 16 + 3 16 ) = 3 8 ≈ 2.67 .
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