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Find the surface area generated by revolving f(x)=x\sqrt{x}x about the x-axis over the interval [1, 4]
34.3
38.14
30.8
What is the formula for area of a surface of revolution is if f(x) is a smooth and non-negative function in the interval [b,c], then the surface area S generated by revolving the curve x=g(y) about the y-axis?
∫cb2\int_c^b2∫cb2 πg(y)1+(dydx)2dx\sqrt{1+\left(\frac{dy}{dx}\right)^2}^{ }dx1+(dxdy)2dx
∫bc\int_b^c∫bc 2 πg(y)1+(dxdy)2dy\sqrt{1+\left(\frac{dx}{dy}\right)^2}^{ }dy1+(dydx)2dy
∫ab\int_a^b∫ab 2 πg(x)1+(dydx)2dx\sqrt{1+\left(\frac{dy}{dx}\right)^2}^{ }dx1+(dxdy)2dx
Calculate area of a surface of revolution generated by revolving semi-circle of radius r=0.5m about the x-axis
2π m2
4 m2
4π m2
Find the surface area that is generated by revolving y=x3 on [0, 2] about the x-axis
205
1203
203
Find the surface area generated by revolving f(y) =13y3\frac{1}{3}y^331y3 about the y-axis over the interval [0, 2]
24.11
13.13
10.8
Find the area of the surface generated by revolving the graph of f(x)= x2 on the interval [0,3\sqrt{3}3 ] about the y-axis
34
23
45
Find the surface area generated by revolving f(y)=9−y2\sqrt{9-y^2}9−y2 about the y-axis over the interval [0, 2]
13 π
12π
Find the surface area generated by revolving f(x)=1−x\sqrt{1-x}1−x about the x-axis over the interval [0, 1/2]
4.3
3.13
Find the surface area generated by revolving f(x)= e0.1x about the x-axis over the interval [0, 4]
31.14
Calculate area of a surface of revolution generated by revolving semi-circle of radius r about the x-axis
4πr
4πr3
4πr2
It is done.