GATE CSE 2023 | Question: 21

Question

The value of the definite integral
\[
\int_{-3}^{3} \int_{-2}^{2} \int_{-1}^{1}\left(4 x^{2} y-z^{3}\right) \mathrm{d} z \mathrm{~d} y \mathrm{~d} x
\]
is _________. (Rounded off to the nearest integer)

Comments

The correct answer is  0

---

0

---

The function is odd so integral will be 0

Answer 1

$\int_{-3}^{3} \int_{-2}^{2} \int_{-1}^{1}\left(4 x^{2} y-z^{3}\right) \mathrm{d} z \mathrm{~d} y \mathrm{~d} x$

= $\int_{-3}^{3} \int_{-2}^{2} \left[4 x^{2} y z – \frac{z^{4}}{4}\right]_{-1}^1 \mathrm{~d} y \mathrm{~d} x$

= $\int_{-3}^{3} \int_{-2}^{2}\left(8 x^{2} y \right) \mathrm{~d} y \mathrm{~d} x$

= $\int_{-3}^{3} \left[4 x^{2} y^2 \right]_{-2}^{2}  \mathrm{~d} x$

= 0

Answer is 0

Answer 2

$\int_{-a}^{a}\int_{-b}^{b}\int_{-c}^{c} f(x,y,z) \ dz \ dy \ dx = 0$ if $f(-x,-y,-z) = -f(x,y,z)$ (a,b and c are constants)

Though the given integral is not hard to compute but still you can solve this question within some seconds.

Here, $f(x,y,z) = 4x^2y \ – z^3$

and here $f(-x,-y,-z) = -f(x,y,z)$

Therefore, answer is zero.

---

Edit:

Some points can be noted as:

$1.$ The above property helps to solve various kind of problems where integrand looks difficult to integrate.

$2.$ This property is an extended form of $\int_{-a}^{a}f(x)dx=0$ when $f(-x)=-f(x)$ i.e. an odd function. It is also applicable in integral with two or more than two variables but here we use the term odd-symmetric function instead of odd function. The proof is also easy and similar which is given at the end.

$3.$ The order of integration does not matter but we have to take care of limits.

For example, $\int_{z=0}^{z=3}\int_{y=0}^{y=2}\int_{x=0}^{x=1}xy^2z \;\;dxdydz=\int_{z=0}^{z=3}\int_{x=0}^{x=1}\int_{y=0}^{y=2}xy^2z \;\;dydxdz$

$4.$  The R code can also be written  to evaluate the given integral.

 

library(cubature)
f <- function(x) { 4 * (x[1]^2*x[2]) - x[3]^3 } # "x" is vector
print(adaptIntegrate(f, lowerLimit = c(-3, -2, -1), upperLimit = c(3, 2, 1))$integral)

----------------------------------------------------------------------------------------------------------------

Proof of the given property:

$I = \int_{-a}^{a}\int_{-b}^{b}\int_{-c}^{c} f(x,y,z) \ dz \ dy \ dx $

$ = \int_{x= -a}^{0}\int_{y = -b}^{0}\int_{z = -c}^{0} f(x,y,z) \ dz \ dy \ dx + \int_{x= 0}^{a}\int_{y = 0}^{b}\int_{z = 0}^{c} f(x,y,z) \ dz \ dy \ dx$

Say, $I = P + Q$

So, $Q= \int_{x= 0}^{a}\int_{y = 0}^{b}\int_{z = 0}^{c} f(x,y,z) \ dz \ dy \ dx$

Let, $x = -u, y = -v, z = -w$

$Q = (-1)^3 \int_{u= 0}^{-a}\int_{v = 0}^{-b}\int_{w = 0}^{-c} f(-u,-v,-w) \ dw \ dv \ du$

$Q = - \int_{u= 0}^{-a}\int_{v = 0}^{-b}\int_{w = 0}^{-c} -f(u,v,w) \ dw \ dv \ du$

$Q =  \int_{u= 0}^{-a}\int_{v = 0}^{-b}\int_{w = 0}^{-c} f(u,v,w) \ dw \ dv \ du$

$Q = (-1)^3 \int_{u= -a}^{0}\int_{v = -b}^{0}\int_{w = -c}^{0} f(u,v,w) \ dw \ dv \ du$

$Q = - \int_{x= -a}^{0}\int_{y = -b}^{0}\int_{z = -c}^{0} f(x,y,z) \ dz \ dy \ dx$

$Q = -P$

Hence, $I = P + Q = 0$