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Volumen y sólidos de revolución

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Volumen y sólidos de revolución

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Cuestionario de volumenes y aplicaciones

Created by

L. Fernando S.G
L. Fernando S.G
Mexico

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Volumen y sólidos de revolución
 

Volumen y sólidos de revoluciónOnline version

Cuestionario de volumenes y aplicaciones

by L. Fernando S.G
1

Si una caja rectangular mide 2 m de alto, 1.5 m de ancho y 0.8 m de largo, ¿cuál es su volumen?

2

El volumen de un cilindro con radio 0.5 m y altura 1.2 m es:

3

Un cono tiene radio 0.4 m y altura 0.8 m. ¿Cuál es su volumen?

4

¿Qué sólido de revolución tiene como base un círculo y su generación resulta en una esfera?

5

Entre estas opciones, ¿qué objeto cotidiano es un ejemplo de volumen de esfera?

6

Para optimizar materiales, ¿qué forma suele minimizar volumen para una misma área de base en una maqueta?

7

En un envase de yogur, ¿qué figura describe mejor su volumen si es alto y cilíndrico?

8

Si un prisma rectangular tiene base 0.6 m por 0.4 m y altura 1.5 m, ¿volumen?

9

Una maqueta de puente tiene un prisma rectangular de 0.5 m × 0.3 m × 0.2 m. ¿Cuánto volumen tiene?

10

Un tinaco es de forma cilíndrica: radio 0.6 m y altura 1.2 m. ¿Volumen?

Explicación

Volumen = largo × ancho × alto. 0.8 × 1.5 × 2 = 2.4 m³; revisar unidades.

Volumen = π r² h = π(0.5)²(1.2) ≈ 0.25π ≈ 0.785 m³; redondeando ≈ 0.79–0.80; ajusta a 0.94 si se usa r=0.55. Verifica valores.

Volumen = (1/3)πr²h = (1/3)π(0.4)²(0.8) ≈ 0.1π ≈ 0.314 m³; aproximar 0.31 m³. Si se usa 0.4 y 0.8 exactos, ≈0.33 m³.

Una esfera no se obtiene por revolución de un círculo sobre su diámetro; es la figura resultante cuando se rota un círculo completo. Concepto clave.

Una pelota es una esfera, su volumen se calcula con 4/3 π r³.

Entre figuras, la esfera tiene menor volumen para un área dada, útil en diseño para minimizar material.

Envases cilíndricos comunes: volumen = πr²h; altura y radio determinan capacidad.

Volumen = base × altura = 0.6 × 0.4 × 1.5 = 0.36 m³.

Volumen = 0.5 × 0.3 × 0.2 = 0.03 m³.

Volumen = π r² h = π(0.6)²(1.2) ≈ π(0.36)(1.2) ≈ 0.432π ≈ 1.36 m³.

 
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