anonymous
  • anonymous
write the definite integral that represents the area of the surface formed by revolving the graph f(x)=x^(1/2) on the interval [0,4] about the y-axis (Do not evaluate the integral.)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
can you help me with the steps to get to that answer? i don't understand how you got there
anonymous
  • anonymous
The surface area can be considered the sum of the circumferences of an infinite number of circles:|dw:1371174218727:dw| The circumference of one such circle is \(2\pi r\), where \(r\) is the radius of each circle. The integral would then look something like this: \[A=2\pi\int_a^b(\text{radius})~ds\] Here, \(ds=\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}~dx\). I'm not as familiar with the derivation for \(ds\), but this link should explain it: http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx The radius of each circle is the distance from the curve to the axis of revolution. In this case, the axis is the y-axis, or \(x=0\). The distance from any given point on the curve to this axis is thus \(x\).
anonymous
  • anonymous
\[\bf SA = \int\limits_{0}^{4} 2\pi x \sqrt{1+\left( \frac{ 1 }{ 2 \sqrt{x} } \right)^2}dx\]

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