Applications of definite integral

Volume of a Solid

Submitted by petr.beremlijski on Tue, 10/22/2024 - 13:41

Area Under the Curve I

Submitted by michaela.bailova on Wed, 10/16/2024 - 11:33

2010014706

Level:

In the experimental process, the ideal gas is expanded adiabatically from an initial volume of

V_{1} = 0.3 m^{3}

to a final volume of

V_{2} = 0.8 m^{3}

. Find a work done by the gas in the given process. Hint: An adiabatic process with an ideal gas follows the relationship

p V^{1.4} = c

, where

p

is a gas pressure,

V

is a gas volume, and

c

is a positive constant. The work

W

done by a gas is defined as

W = \int_{V_{1}}^{V_{2}} p d V

W ≐ 1.313 c J

W ≐ 0.375 c J

W ≐ 6.782 c J

W ≐ 0.221 c J

2010014705

Level:

In the experimental process, the ideal gas is expanded isothermally from an initial pressure of

0.8 MPa

and volume of

V_{1} = 0.3 m^{3}

to a final volume of

V_{2} = 1.2 m^{3}

. Find a work done by the gas in the given process. Hint: During isothermal expansion, both pressure

p

and volume

V

change along an isotherm with a constant

p V

product. The work

W

done by a gas is defined as

W = \int_{V_{1}}^{V_{2}} p d V

W ≐ 333 kJ

W ≐ 216 kJ

W ≐ 720 kJ

W ≐ 178 kJ

2010014704

Level:

The time characteristic of an alternating current

i

is given in the figure. Find the effective value

I

of the alternating current

i

provided the following relation holds:

I^{2} T = \int_{0}^{T} i^{2} d t

I = 500 mA

I = 354 mA

I = 0 mA

I = 250 mA

2010014703

Level:

The time characteristic of an alternating voltage

u

is given in the figure. Find the effective value

U

of the alternating voltage

u

provided the following relation holds:

U^{2} T = \int_{0}^{T} u^{2} d t

U = 325 V

U ≐ 230 V

U = 0 V

U = \frac{325}{2} V

2010014702

Level:

The time characteristic of an alternating voltage

u

is given in the figure, where

U_{m}

is the peak value of

u

. Find the effective value

U

of the alternating voltage

u

provided the following relation holds:

U^{2} T = \int_{0}^{T} u^{2} d t

U = \frac{\sqrt{3}}{3} U_{m}

U = \frac{\sqrt{2}}{2} U_{m}

U = \frac{1}{3} U_{m}

U = \frac{1}{2} U_{m}

2010014701

Level:

The time characteristic of an alternating current

i

is given in the figure, where

I_{m}

is the peak value of

i

. Find the effective value

I

of the alternating current

i

provided the following relation holds:

I^{2} T = \int_{0}^{T} i^{2} d t

I = \frac{\sqrt{3}}{3} I_{m}

I = \frac{\sqrt{2}}{2} I_{m}

I = \frac{1}{3} I_{m}

I = \frac{1}{2} I_{m}

2010014306

Level:

Approximately, the shape of Mars is an ellipsoid. This ellipsoid can be obtained by rotating an ellipse with semi-axes

a = 3 396 190 m

and

b = 3 376 200 m

around its minor axis. What is the volume

V

of this ellipsoid?

V ≐ 1.631 \cdot 10^{20} m^{3}

V ≐ 1.622 \cdot 10^{20} m^{3}

V ≐ 3.602 \cdot 10^{13} m^{3}

V ≐ 1.132 \cdot 10^{14} m^{3}

2010014305

Level:

Approximately, the shape of the Earth is an ellipsoid. This ellipsoid can be obtained by rotating an ellipse with semi-axes

a = 6 378 137 m

and

b = 6 356 752 m

around its minor axis. What is the volume

V

of this ellipsoid?

V ≐ 1.083 \cdot 10^{21} m^{3}

V ≐ 1.080 \cdot 10^{21} m^{3}

V ≐ 4.002 \cdot 10^{14} m^{3}

V ≐ 1.274 \cdot 10^{14} m^{3}

Applications of definite integral

Volume of a Solid

Area Under the Curve I

2010014706

2010014705

2010014704

2010014703

2010014702

2010014701

2010014306

2010014305

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