The given curve is rotated about the y -axis. Find the area of the resulting surface. y=(1)/(4)

The given curve is rotated about the y -axis. Find the area...

Key Concepts

Definite Integration

Definite integration is used to accumulate the contributions of all the infinitesimal surface elements over a specified interval. In the context of surface area problems, the integration bounds define the portion of the curve being rotated. Mastery of techniques in definite integration allows one to evaluate the overall surface area, particularly when the integrands involve composite functions or radicals arising from the arc length differential.

Arc Length Differential

The arc length differential element, typically expressed as ds = sqrt(1 + (dy/dx)^2) dx when the curve is given in Cartesian form, is fundamental in these types of problems. It quantifies the infinitesimal distance along the curve and, when multiplied by the revolving quantity (such as the distance from the axis of rotation), it yields the small surface area element. Understanding and computing this differential correctly is crucial for setting up the precise integral for the surface area.

Surface Area of Revolution

This concept involves finding the area of a surface generated when a curve is rotated about an axis. The technique relies on integrating infinitesimally small surface elements along the curve, where each element contributes an annular (ring-like) area when revolved around the axis. The general formula incorporates both the radius of rotation and the differential arc length, linking classical calculus integration with geometric visualization.

Transcript

0:00 Great.

00:02 So the given function is 4 x squared minus 1 over 2 times log of x.

00:15 And x is in between these intervals.

00:24 Okay, so the surface area, we're trying to find the surface area of this function, rotated a about the y -axis, and the formula is given by 2 -5 times integral from a to b, x of y times square root 1 plus x prime of y square with respect to y.

01:02 Okay, so the first thing that we want to do, because this function cannot be cannot let us find the expression of x explicitly in terms of y so we we need to find the approach that doesn't involve finding x in terms of y explicitly so all we're going to do is to use the u substitution let's set u is equal to x of y that means the u d y is equal to x of y d u d y is equal to x prime of y and that means 1 over x y d u d u d u is equal to x prime of y d u is equal to so after substituting this expression into the into our integral equation we have okay the the thing that i want to mention was now that u is equal to x x of y the the lower limit and upper limit this is just one to two.

02:58 Because u is equal to x of y okay so now we do this one over x prime of y d y not d u okay which can be further simplified to one plus one over x prime of y square d u okay and now we have to find the expression of x prime of y in terms of u or in terms of x because these are the same thing.

03:54 From the function given, we can find the expression of x prime of y just by taking the derivative of this guy with respect to y.

04:18 Okay, so let's do that.

04:22 D o d y 1 over 4 x square minus 1 over 2 and x the left side is just 1 and the right side becomes 1 of 4 2 x times x minus 1 over 2 times 1 of x times x prime.

04:57 So the, i should have written x of y because we're taking the derivative of this expression with respect y and dash should have clarified more, but you get the idea times x prime...

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