What so the equation that passes through the point (-1,1) and has a sleop of 1/6?
1 answer:
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Write the given equation as
x = (1/2)y² or as y = √(2x)
Graph the given curve within the region (0,0) and (2,2) as shown in the figure below.
When the curve is rotated about the x-axis, an element of surface area is
dA = 2πy dx
The surface area of the resulting solid is
![A= 2\pi \int_{0}^{2} \sqrt{2x} dx = \frac{4 \sqrt{2} \pi}{3} [x^{3/2}]_{0}^{2} = \frac{16 \pi}{3}](https://tex.z-dn.net/?f=A%3D%202%5Cpi%20%5Cint_%7B0%7D%5E%7B2%7D%20%20%5Csqrt%7B2x%7D%20dx%20%3D%20%20%5Cfrac%7B4%20%5Csqrt%7B2%7D%20%5Cpi%7D%7B3%7D%20%5Bx%5E%7B3%2F2%7D%5D_%7B0%7D%5E%7B2%7D%20%3D%20%5Cfrac%7B16%20%5Cpi%7D%7B3%7D%20)
If the right end is considered, the extra area is π*(2²) = 4π
Answer:
The surface area of the rotated solid is (16π)/3.
If the right end is considered, it is an extra area of 4π.
x = (1/2)y² or as y = √(2x)
Graph the given curve within the region (0,0) and (2,2) as shown in the figure below.
When the curve is rotated about the x-axis, an element of surface area is
dA = 2πy dx
The surface area of the resulting solid is
If the right end is considered, the extra area is π*(2²) = 4π
Answer:
The surface area of the rotated solid is (16π)/3.
If the right end is considered, it is an extra area of 4π.