Example Question - calculating volumes
Here are examples of questions we've helped users solve.
Calculating Volumes using the Disk Method for Solid of Revolution
Para resolver este problema, utilizaremos el método de los discos para calcular el volumen de una figura de revolución. El segmento de recta dado por la ecuación \( y = 2x + 1 \) se gira alrededor del eje x para formar un volumen tridimensional, y queremos calcular el volumen de este sólido entre \( x = 1 \) y \( x = 5 \). La fórmula para el volumen \( V \) del sólido de revolución generado al girar una función \( y = f(x) \) alrededor del eje x, entre \( x = a \) y \( x = b \), es: \[ V = \pi \int_{a}^{b} [f(x)]^2 dx \] Dado que \( f(x) = 2x + 1 \), la integral que necesitamos resolver es: \[ V = \pi \int_{1}^{5} (2x + 1)^2 dx \] Desarrollamos el cuadrado de la función dentro de la integral: \[ (2x + 1)^2 = 4x^2 + 4x + 1 \] Ahora integramos término por término: \[ V = \pi \int_{1}^{5} (4x^2 + 4x + 1) dx \] \[ V = \pi \left[ \frac{4x^3}{3} + 2x^2 + x \right]_{1}^{5} \] Evaluamos la integral definida: \[ V = \pi \left( \left[ \frac{4(5^3)}{3} + 2(5^2) + 5 \right] - \left[ \frac{4(1^3)}{3} + 2(1^2) + 1 \right] \right) \] \[ V = \pi \left( \left[ \frac{4 \cdot 125}{3} + 50 + 5 \right] - \left[ \frac{4}{3} + 2 + 1 \right] \right) \] \[ V = \pi \left( \left[ \frac{500}{3} + 55 \right] - \left[ \frac{7}{3} \right] \right) \] \[ V = \pi \left( \frac{500}{3} + 55 - \frac{7}{3} \right) \] \[ V = \pi \left( \frac{500 + 165 - 7}{3} \right) \] \[ V = \pi \left( \frac{658}{3} \right) \] Por lo tanto, el volumen del cono generado es: \[ V = \frac{658\pi}{3} \text{ unidades cúbicas} \]
Calculating Volumes and Areas
Let's solve both parts of this question one by one: ### Part a) To calculate the volume of the pyramid with a square base: The formula for the volume of a pyramid is \( \frac{1}{3} \times \text{base area} \times \text{height} \). Given: - The height (\(h\)) of the pyramid is 10 cm. - The side (\(s\)) of the square base is 5 cm. First, calculate the base area: - Base area = \( s^2 = 5 \text{ cm} \times 5 \text{ cm} = 25 \text{ cm}^2 \). Next, calculate the volume of the pyramid: - Volume = \( \frac{1}{3} \times \text{base area} \times h = \frac{1}{3} \times 25 \text{ cm}^2 \times 10 \text{ cm} \). - Volume = \( \frac{1}{3} \times 25 \times 10 \text{ cm}^3 = \frac{250}{3} \text{ cm}^3 \approx 83.33 \text{ cm}^3 \). To convert cubic centimeters to liters, remember that 1 liter equals 1000 cubic centimeters: - Volume in liters = \( \frac{83.33}{1000} \text{ liters} \approx 0.083 \text{ liters} \). So the volume of the pyramid is approximately 0.083 liters (to 2 decimal places: 0.08 liters). ### Part b) For the rectangular cover plate with rounded corners to form sectors: Given: - The dimensions of the rectangular plate are 500 mm by 300 mm. - Each corner is rounded to form a sector with a radius of 25 mm. Firstly, calculate the area of the original rectangle: - Area of rectangle = Length × Width = 500 mm × 300 mm = 150000 mm². Next, you have to subtract the area of the four sectors. Each sector is a quarter of a circle with a radius of 25 mm. The area of one full circle with radius \(r\) is \( \pi r^2 \). Therefore, the area of one quarter-circle (or sector) would be \( \frac{1}{4} \pi r^2 \). - Area of one sector = \( \frac{1}{4} \pi (25 \text{ mm})^2 \). - Area of one sector = \( \frac{1}{4} \pi \times 625 \text{ mm}^2 \). - Area of one sector = \( 156.25 \pi \text{ mm}^2 \). Since there are four such sectors: - Total area of the four sectors = \( 4 \times 156.25 \pi \text{ mm}^2 \). - Total area of the four sectors = \( 625 \pi \text{ mm}^2 \). Subtracting this from the area of the rectangle gives the final area: - Final area of the plate = Area of rectangle - Total area of the four sectors. - Final area of the plate = 150000 mm² - 625π mm². - Final area of the plate = \( 150000 \text{ mm}^2 - 1963.4954 \text{ mm}^2 \) (using \( \pi \approx 3.1416 \)). - Final area of the plate ≈ 147036.5 mm² (rounded to one decimal place). So, the final area of the rectangular cover plate, after rounding off the corners to form sectors, is approximately 147036.5 mm².
Calculating Volumes of Rectangular Prisms and Polyhedron
To calculate the volume of each rectangular prism and the combined polyhedron, use the formula: Volume of a rectangular prism = length × width × height For Rectangular Prism 1: - Length = 11 mm - Width = 5 mm - Height = 8 mm Volume = 11 mm × 5 mm × 8 mm = 440 cubic millimeters For Rectangular Prism 2: - Length = 15 mm - Width = 8 mm - Height = 8 mm Volume = 15 mm × 8 mm × 8 mm = 960 cubic millimeters Now for the combined polyhedron which is the sum of the volumes of the two prisms: Volume = Volume of Prism 1 + Volume of Prism 2 Volume = 440 mm³ + 960 mm³ = 1400 mm³ So, the answers are: A) The volume of Rectangular Prism 1 is 440 cubic millimeters. B) The volume of Rectangular Prism 2 is 960 cubic millimeters. C) The volume of the polyhedron is 1400 cubic millimeters.
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