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Calculus

GATE CSE 2023 · Q31

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The value of the definite integral ∫−33∫−22∫−11(4x2y−z3) dz dy dx\int_{-3}^{3} \int_{-2}^{2} \int_{-1}^{1} (4x^2y - z^3) \, dz \, dy \, dx is _________ . (Rounded off to the nearest integer)

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Calculus

We are asked to evaluate the triple integral:

I=∫−33∫−22∫−11(4x2y−z3) dz dy dxI = \int_{-3}^{3} \int_{-2}^{2} \int_{-1}^{1} (4x^2y - z^3) \, dz \, dy \, dx

By linearity, the integral can be split into two separate integrals:

I=I1−I2I = I_1 - I_2

where:

I1=∫−33∫−22∫−114x2y dz dy dxI_1 = \int_{-3}^{3} \int_{-2}^{2} \int_{-1}^{1} 4x^2y \, dz \, dy \, dx
I2=∫−33∫−22∫−11z3 dz dy dxI_2 = \int_{-3}^{3} \int_{-2}^{2} \int_{-1}^{1} z^3 \, dz \, dy \, dx

Since the integration limits are independent constants, each multiple integral can be factored into single variable integrals:

I1=4(∫−33x2 dx)(∫−22y dy)(∫−111 dz)I_1 = 4 \left( \int_{-3}^{3} x^2 \, dx \right) \left( \int_{-2}^{2} y \, dy \right) \left( \int_{-1}^{1} 1 \, dz \right)

The function yy is an odd function, and the limits of integration [−2,2][-2, 2] are symmetric about zero:

∫−22y dy=[y22]−22=42−42=0\int_{-2}^{2} y \, dy = \left[ \frac{y^2}{2} \right]_{-2}^{2} = \frac{4}{2} - \frac{4}{2} = 0

Thus, I1=0I_1 = 0.

Similarly, for I2I_2:

I2=(∫−331 dx)(∫−221 dy)(∫−11z3 dz)I_2 = \left( \int_{-3}^{3} 1 \, dx \right) \left( \int_{-2}^{2} 1 \, dy \right) \left( \int_{-1}^{1} z^3 \, dz \right)

Since z3z^3 is also an odd function over the symmetric interval [−1,1][-1, 1]:

∫−11z3 dz=[z44]−11=14−14=0\int_{-1}^{1} z^3 \, dz = \left[ \frac{z^4}{4} \right]_{-1}^{1} = \frac{1}{4} - \frac{1}{4} = 0

Thus, I2=0I_2 = 0.

Therefore:

I=I1−I2=0−0=0I = I_1 - I_2 = 0 - 0 = 0

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This Calculus previous year question is Q31 of the GATE CSE 2023 paper. Find more GATE CSE PYQs on Calculus in Engineering Mathematics, or browse all GATE CSE PYQs.

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The value of the definite integral -33 -22 -11 (4x2y - z3) dz dy dx is … (Rounded off to the nearest integer)

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