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Math  /  Calculus

Question1. [-/7.14 Points]
DETAILS MY NOTES SCALCET9 6.3.057.
The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method. x2+(y−4)2=16;x^{2}+(y-4)^{2}=16 ; about the yy-axis □\square Need Help? Read It Watch it

Studdy Solution

STEP 1

1. The given curve is a circle with the equation x2+(y−4)2=16x^2 + (y-4)^2 = 16.
2. The circle is centered at (0,4)(0, 4) with a radius of 44.
3. The rotation is about the yy-axis.
4. The volume of the resulting solid can be found using the method of cylindrical shells or the disk/washer method.

STEP 2

1. Rewrite the equation of the circle in a form convenient for integration.
2. Set up the integral for the volume using the disk/washer method or cylindrical shells.
3. Evaluate the integral to find the volume.

STEP 3

Rewrite the equation x2+(y−4)2=16x^2 + (y-4)^2 = 16 to express xx in terms of yy. x2=16−(y−4)2 x^2 = 16 - (y-4)^2 x=16−(y−4)2andx=−16−(y−4)2 x = \sqrt{16 - (y-4)^2} \quad \text{and} \quad x = -\sqrt{16 - (y-4)^2}

STEP 4

Set up the integral for the volume using the disk/washer method. The volume VV of the solid of revolution about the yy-axis can be expressed as: V=π∫y1y2[R(y)2−r(y)2] dy V = \pi \int_{y_1}^{y_2} \left[ R(y)^2 - r(y)^2 \right] \, dy Here, R(y)R(y) is the outer radius and r(y)r(y) is the inner radius. Since the circle is symmetric and centered at (0,4)(0, 4), there is no inner radius.

STEP 5

Determine the limits of integration, y1y_1 and y2y_2. The circle extends from y=0y=0 to y=8y=8. y1=0,y2=8 y_1 = 0, \quad y_2 = 8

STEP 6

In this case, the outer radius R(y)R(y) is x=16−(y−4)2x = \sqrt{16 - (y-4)^2} and the inner radius r(y)=0r(y) = 0. V=π∫08[16−(y−4)2]2 dy V = \pi \int_{0}^{8} \left[ \sqrt{16 - (y-4)^2} \right]^2 \, dy

STEP 7

Simplify the integral. V=π∫08[16−(y−4)2] dy V = \pi \int_{0}^{8} \left[ 16 - (y-4)^2 \right] \, dy

STEP 8

Evaluate the integral. V=π∫08[16−(y−4)2] dy V = \pi \int_{0}^{8} \left[ 16 - (y-4)^2 \right] \, dy Separate the integral into two parts: V=π(∫0816 dy−∫08(y−4)2 dy) V = \pi \left( \int_{0}^{8} 16 \, dy - \int_{0}^{8} (y-4)^2 \, dy \right)

STEP 9

Evaluate the first integral: ∫0816 dy=16y∣08=16(8)−16(0)=128 \int_{0}^{8} 16 \, dy = 16y \Big|_{0}^{8} = 16(8) - 16(0) = 128

STEP 10

Evaluate the second integral. First, let u=y−4u = y - 4: ∫08(y−4)2 dy=∫−44u2 du \int_{0}^{8} (y-4)^2 \, dy = \int_{-4}^{4} u^2 \, du

STEP 11

Evaluate the integral ∫−44u2 du\int_{-4}^{4} u^2 \, du: ∫−44u2 du=u33∣−44=(4)33−(−4)33=643−(−643)=1283 \int_{-4}^{4} u^2 \, du = \frac{u^3}{3} \Big|_{-4}^{4} = \frac{(4)^3}{3} - \frac{(-4)^3}{3} = \frac{64}{3} - \left(-\frac{64}{3}\right) = \frac{128}{3}

STEP 12

Combine the results of the integrals: V=π(128−1283)=π(3843−1283)=π(2563)=256π3 V = \pi \left( 128 - \frac{128}{3} \right) = \pi \left( \frac{384}{3} - \frac{128}{3} \right) = \pi \left( \frac{256}{3} \right) = \frac{256\pi}{3}
Solution: The volume of the resulting solid is: V=256π3 V = \frac{256\pi}{3}

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