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Pregunta 6 Evalúa la integral que representa el volumen de un cilindro de radio R y altura \( h \) en coordenadas cilíndricas: \( \int_{0}^{2 \pi} \int_{0}^{R} \int_{0}^{h} r d z d r d \theta \)

Ask by Hamilton Munoz. in Mexico
Mar 10,2025

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Answer

El volumen del cilindro es \( \pi R^2 h \).

Solution

1. Se evalúa primero la integral respecto a \( z \): \[ \int_{0}^{h} dz = h. \] 2. La integral se simplifica a: \[ h \int_{0}^{2 \pi} \int_{0}^{R} r \, dr \, d\theta. \] 3. Se evalúa la integral respecto a \( r \): \[ \int_{0}^{R} r \, dr = \frac{R^2}{2}. \] 4. La integral se reduce a: \[ h \cdot \frac{R^2}{2} \int_{0}^{2 \pi} d\theta. \] 5. Se evalúa la integral respecto a \( \theta \): \[ \int_{0}^{2 \pi} d\theta = 2\pi. \] 6. Sustituyendo los resultados obtenemos: \[ h \cdot \frac{R^2}{2} \cdot 2\pi = \pi R^2 h. \] Por lo tanto, el volumen del cilindro es: \[ \pi R^2 h. \]

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Para calcular el volumen del cilindro utilizando las coordenadas cilíndricas, comenzamos evaluando la integral: \[ \int_{0}^{2 \pi} \int_{0}^{R} \int_{0}^{h} r \, d z \, d r \, d \theta. \] Primero, integramos respecto a \( z \): \[ \int_{0}^{h} dz = h. \] Reemplazamos esto en la integral: \[ \int_{0}^{2 \pi} \int_{0}^{R} r \cdot h \, d r \, d \theta = h \int_{0}^{2 \pi} d \theta \int_{0}^{R} r \, d r. \] Ahora, evaluamos la integral sobre \( r \): \[ \int_{0}^{R} r \, d r = \frac{R^2}{2}. \] Sustituyendo, tenemos: \[ h \int_{0}^{2 \pi} d \theta \cdot \frac{R^2}{2} = h \cdot 2\pi \cdot \frac{R^2}{2} = \pi R^2 h. \] Así que el volumen \( V \) del cilindro es: \[ V = \pi R^2 h. \]

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