Pregunta
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Let \( R \) be the region bounded by the following curves. Use the method of your choice to find the volume of the solid generated when \( R \) is revolved about the \( x \)-axis. \( y=4 x^{2} \) and \( y=20-x^{2} \) The volume of the solid is (Type an exact answer.)

Ask by Byrd Nguyen. in the United States
Dec 08,2024

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The volume of the solid is \( 1088\pi \).

Solución

To find the volume of the solid generated when the region \( R \) bounded by the curves \( y = 4x^2 \) and \( y = 20 - x^2 \) is revolved about the \( x \)-axis, we can use the disk method. First, we need to find the points of intersection of the curves \( y = 4x^2 \) and \( y = 20 - x^2 \). We set the equations equal to each other: \[ 4x^2 = 20 - x^2 \] Rearranging gives: \[ 4x^2 + x^2 = 20 \] \[ 5x^2 = 20 \] \[ x^2 = 4 \] \[ x = \pm 2 \] Thus, the curves intersect at the points \( x = -2 \) and \( x = 2 \). Next, we determine the volume of the solid of revolution using the disk method. The volume \( V \) is given by the integral: \[ V = \pi \int_{a}^{b} (R(x)^2 - r(x)^2) \, dx \] where \( R(x) \) is the outer radius and \( r(x) \) is the inner radius. In this case, the outer radius is given by the upper curve \( y = 20 - x^2 \) and the inner radius is given by the lower curve \( y = 4x^2 \). Thus, we have: \[ R(x) = 20 - x^2 \] \[ r(x) = 4x^2 \] The volume integral becomes: \[ V = \pi \int_{-2}^{2} \left((20 - x^2)^2 - (4x^2)^2\right) \, dx \] Now we compute \( (20 - x^2)^2 \) and \( (4x^2)^2 \): \[ (20 - x^2)^2 = 400 - 40x^2 + x^4 \] \[ (4x^2)^2 = 16x^4 \] Substituting these into the volume integral gives: \[ V = \pi \int_{-2}^{2} \left(400 - 40x^2 + x^4 - 16x^4\right) \, dx \] \[ = \pi \int_{-2}^{2} \left(400 - 40x^2 - 15x^4\right) \, dx \] Now we can evaluate the integral: \[ V = \pi \left[ 400x - 20x^3 - 3x^5 \right]_{-2}^{2} \] Calculating at the bounds: At \( x = 2 \): \[ 400(2) - 20(2^3) - 3(2^5) = 800 - 20(8) - 3(32) = 800 - 160 - 96 = 544 \] At \( x = -2 \): \[ 400(-2) - 20(-2)^3 - 3(-2)^5 = -800 + 20(8) + 3(32) = -800 + 160 + 96 = -544 \] Thus, the definite integral evaluates to: \[ V = \pi \left( 544 - (-544) \right) = \pi (544 + 544) = \pi (1088) \] Therefore, the volume of the solid is: \[ \boxed{1088\pi} \]

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The Deep Dive

To find the volume of the solid generated by revolving the region \( R \) bounded by the curves \( y = 4x^2 \) and \( y = 20 - x^2 \) about the \( x \)-axis, we first need to find the points of intersection of the two curves. Setting \( 4x^2 = 20 - x^2 \): \[ 4x^2 + x^2 = 20 \\ 5x^2 = 20 \\ x^2 = 4 \\ x = \pm 2 \] The points of intersection are \( x = -2 \) and \( x = 2 \). Next, we determine the volume \( V \) using the washer method: \[ V = \pi \int_{-2}^{2} \left( (20 - x^2)^2 - (4x^2)^2 \right) \, dx \] Calculating \( (20 - x^2)^2 \) and \( (4x^2)^2 \): \[ (20 - x^2) = 400 - 40x^2 + x^4 \] \[ (4x^2)^2 = 16x^4 \] Now substituting these values in the volume integral: \[ V = \pi \int_{-2}^{2} \left( (400 - 40x^2 + x^4) - (16x^4) \right) \, dx \] \[ = \pi \int_{-2}^{2} (400 - 40x^2 - 15x^4) \, dx \] Because the integrand is even, we can simplify the volume calculation: \[ V = 2\pi \int_{0}^{2} (400 - 40x^2 - 15x^4) \, dx \] Calculating the integral: \[ \int (400 - 40x^2 - 15x^4) \, dx = 400x - \frac{40}{3}x^3 - 3x^5/5 \] Evaluating from 0 to 2: \[ = \left[ 400(2) - \frac{40}{3}(2^3) - 3\frac{(2^5)}{5} \right] - \left[ 400(0) - \frac{40}{3}(0^3) - 3\frac{(0^5)}{5} \right] \] \[ = 800 - \frac{40}{3}(8) - \frac{3}{5}(32) \] \[ = 800 - \frac{320}{3} - \frac{96}{5} \] Finding a common denominator (15): \[ = 800 - \frac{1600}{15} - \frac{288}{15} \\ = 800 - \frac{1888}{15} \\ = \frac{12000}{15} - \frac{1888}{15} \\ = \frac{10112}{15} \] Now multiplying by \( 2\pi \): \[ V = 2\pi \cdot \frac{10112}{15} = \frac{20224\pi}{15} \] Thus, the volume of the solid is: \[ \boxed{\frac{20224\pi}{15}} \]

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