Compute the surface area of revolution about the x -axis over the interval. y=(4-x^2 / 3)^3

Compute the surface area of revolution about the x -axis...

Key Concepts

Surface Area of Revolution

This concept involves calculating the area of a surface generated when a curve is revolved about an axis. The formula for revolution about the x-axis, for example, is given by SA = 2? ?[a, b] f(x) ?(1 + (f'(x))²) dx. It requires setting up the integral with the appropriate function and its derivative, followed by evaluating the integral over the given interval.

Chain Rule in Differentiation

The chain rule is essential for finding the derivative of composite functions, especially when the function involves nested functions or fractional exponents. It allows us to differentiate a function like y = (4 - x^(2/3))^(3/2) by differentiating the outer function and multiplying by the derivative of the inner function.

Integration Techniques

Computing the surface area of revolution often leads to integrals that are nontrivial. Techniques such as u-substitution or trigonometric substitution are used to simplify the integrand. Mastery of these methods is crucial in handling the expressions that arise from the derivative and the square root term in the surface area formula.

Algebraic Manipulation and Simplification

Before integrating, it is often necessary to simplify complex expressions involving radicals and fractional exponents. This step makes the integrals more tractable by reducing them to a form where standard integration techniques can be applied effectively.

Transcript

00:01 In this problem, we are going to compute the surface area of this function from 0 to 8 when it's revolved about the x -axis.

00:09 And when we revolve it, we get a figure that looks something like this to the right of our graph.

00:17 And we know that surface area is the integral, in this case from 0 to 8, of 2 pi y, which is a quantity 4, minus.

00:36 X to the two -thirds power all raised to the power of three over two and then of course we have to multiply by the square root of one plus y prime squared let's compute y prime down here if this part is our y then we know that y prime is equal to three halves times the quantity four minus x to the power of 2 over 3 times the quantity we have to subtract 1 so 3 halves minus 1 half minus 3 halves minus 1 which is 1 half and then we have to multiply by the derivative of the inside so we're multiplying by minus 2 thirds times x to the minus 1 3 1 3⁄2s which is equal to 3 halves and 2 thirds they cancel out so we get that this is minus x to the minus one -third times a quantity four minus x to the two -thirds all to the power of one -half and so now when we square y prime we get that this is equal to x to the minus two -thirds times the quantity four minus x to the two -thirds so we can write that um under this radical, so this is 1 plus x to the minus 2 thirds, times the quantity 4 minus x to the 2 thirds, dx.

03:18 And so we get that this is the integral from 0 to 8 of 2 pi.

03:27 We still have to multiply by our function here, and that's times the square root of 1.

03:46 Here we can distribute x to the minus two -thirds.

03:51 So we get that this is one plus four times x to the minus two -thirds minus one dx, which is equal to the integral of two -pie.

04:20 And then these ones, they cancel each other, one minus one.

04:25 That's just zero.

04:26 So this is two -pie times a square root of four -x to the minus two -thirds, which is just two times x to the minus one -third times our original function.

04:56 And then from here we can use a u -substitution.

05:02 So we can let u be 4 minus x to the two -thirds.

05:11 So that way, du is minus two -thirds times x to the minus one -third d -x...

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