Find the exact area of the surface obtained by rotating the curve about the x-axis. y=(x^3)/(6)+

Find the exact area of the surface obtained by rotating the...

Key Concepts

Surface Area of Revolution

This concept involves finding the area of a surface generated when a curve is rotated about a given axis. The formula for the surface area, particularly when rotating about the x-axis, is S = 2? ?[a to b] f(x) ?(1 + (dy/dx)²) dx, highlighting the need to integrate the circular arcs formed by the rotation along the curve.

Arc Length Element

The arc length element, given by ds = ?(1 + (dy/dx)²) dx, represents the differential element of the curve's length. It is a critical part of the surface area formula as it accounts for the stretching effect that occurs along the curve when it is revolved, ensuring that the contribution from each infinitesimal segment is accurately measured.

Differentiation

Differentiation is employed to find the derivative of the function defining the curve. The derivative, dy/dx, not only informs about the rate of change of the function but is also used in calculating the arc length element, which is essential for the surface area of revolution.

Definite Integration

Definite integration is used to sum up the contributions of all the infinitesimal surface elements over the specified interval. It provides the exact area by evaluating the integral from the lower to the upper bound, in this case, from x = 1/2 to x = 1.

Algebraic Simplification

This concept involves manipulating the integrand algebraically to simplify the expression under the square root and other parts of the integrand. Recognizing patterns such as perfect squares or common factors can significantly ease the integration process and lead to an exact evaluation of the surface area.

Transcript

00:01 Okay, so we need to find the exact surface area of the obtained by rotating the curve about the x -axis, where the curve is given by the function, y equals x -3 over 6 plus 1 over 2x, and the domain is from 1 of 2 to 1.

00:34 Okay, so as we know, the surface area, formula for the surface area is given by a to b, y of x, 1 plus y prime of x, square, dx, when the curve y is rotated about the x axis from a to b.

01:02 Okay, so now we have a equals 1 over 2, and b equals 1 over 2, and b equals 1.

01:12 And then we need to find y prime of x, which is 3 over 6 squared minus 1 over 2 x squared, which is equal to 1 of 2 x squared, 1 of 2 x squared.

01:37 And then i'm going to simplify this a bit more by making it have the common denominator.

01:53 X to the fourth minus 1 over x squared.

01:58 So that is our expression of y prime of x and then we'll square this expression and get one of four times x to the fourth minus 1 square over x to the 4th.

02:18 Okay.

02:21 So we have all we need to substitute these expression into the surface area from 1 over 2 to 1 and for y we have x to the 3 of 6 plus 1 of 2 x times the square root 1 plus 1 of 4 times x to the 4 minus 1 square over over x to the fourth and then d x.

03:05 Okay.

03:07 What i'm going to do is to, i'm going to call this expression, um, a sub i.

03:16 Okay, so we have a sub i as integral from 1 over 2 to 1, x to the 3rd over 6 plus 1 over 2x times 1 plus 1 .4, x the 4th minus 1 square over x the 4th d x what i'm going to do is to simplify the the function inside the integral to make it easier to for us to integrate okay the first thing i'm going to do is to factor out 1 over 2 and then factor out 1 over 4 x to the fourth from the from inside the square root and when we do that we get the new expression 1 over 2 to 1 1 1 2 x to the 3 or over 3 plus 1 of x times 1 over 2 x squared times 4 x to the 4th plus x to the fourth minus 1 squared d x and then we get 1 of 4 1 over 2 to 1 and then i am going to have this expression have the same denominator and then we get x to the 4 plus 3 over 3 x to the 3 times 4 x to the 4th plus x to the 4th minus 1 square d x x okay um so before before i made this term these terms have the common denominator i just eliminated i just multiplied this 1 over x squared to this expression so that's uh that's how you end up with this term over here okay now notice that there's many x the fourth in the expression so that means we should try u substitution u equals x to the fourth and then that means the lower bound becomes one of the sixteen and then upper bound becomes one and the u is equal to four x to the third dx which implies d x is equal to 1 over 4 x to the 3rd d u okay and then we're going to substitute these expressions into our a sub i and then get 1 of 4 times integral from 1 of 16 to 1 u plus 3 over 3x the 3x the 3 times 4 u plus u minus 1 square times 1 of 4 x the 3rd d u okay so record that we set u equals x the 4th and that means x equals plus or minus u to 1 of 4 and we're taking the integral with respect to x and then x is a positive number so recall that the the original integral limits are 1 over 2 and 1 so x has to be positive which means that x equals u to 1 of 4...

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