Determine whether the following statements are true and give a...

Question

Determine whether the following statements are true and give an explanation or counterexample.

c. Let f(x)=12x^2. The area of the surface generated when the graph of f on [−4, 4] is revolved about the x-axis is twice the area of the surface generated when the graph of f on [0, 4] is revolved about the x-axis.

Accepted Answer

Welcome back, everyone. In this problem, let G of X equals 8 X2. Is the area of the surface generated when the graph of G and the interval from -3 to 3 is revolved above the X-axis equal to twice the error generated when G and the interval 03 is revolved above the x-axis. A says yes and B says no. Now what are we trying to figure out here? Well, if we were to do a quick sketch, OK. Of geoic, and we know that Geoffic is AX squared, so it would be a parabola that intersects our origin or has its vertex at the origin, OK. And let's say that we're looking at the area between 0 and 3, the surface area between 0 and 3, compared to the surface area between 0 and the negative 3. What we're essentially trying to find the question we're asking is that is the surface area. Between 3 and 3 equal to 2 or twice the area generated by the surface area on 03. That's what we're trying to find. Now what do we know about the surface area that's generated when a function is revealed about the x-axis? Well, recall that the surface area is going to be equal to 2 pi multiplied by the definite integral between A and B of G of X multiplied by the square root of 1 plus the derivative of g of X squared. I with respect the X. So in this case, if we can solve for the derivative of geopic and plug it into both of our equations for the different limits for our integrals, then we should be able to compare and see if one surface area is twice the other. Now, given GRX. Eals 8 x 2, right, then the derivative of G of X is going to be equal to 16 X. So, First, For our surface area. On the interval, let's call it, well, you know what for first surface area. On the interval from 0 to 3, OK. Then S1 is gonna be equal to 2 pi multiplied by the definite integral from 0 to 3 of 8 X squared multiplied by the square root of 1 + 16 X squad, which would be 256 X squad all with respect to X. On the other hand, for S2, on the interval from -3 to 3. Then the surface area is going to be equal to 2 pi multiplied by the definite integral between -3 and 3 of 8 X squared multiplied by the square root of 1 plus again 16 x 2, 256 X quad with respect to X. No, 8 x square this expression 8 x squared multiplied by the square root of 1 + 256 x quad is an even expression. So what does that mean? Well, since it's even. OK, then remember, or let me make a note here, recall that for even functions. For evil functions. Then the integral between the limits, let's call them negative A and A of F of X with respect to X will be equal to 2 multiplied by the integral between 0 and A of F of X with respect to X. So this would imply then that S2 is going to be equal to 2 multiplied, OK. By this definite sorry 2 multiplied by 2 multiplied by the definite integral between 0 and 3 of our expression X2 8 x2 multiplied by 1 + 256 X2d with respect to X. But remember, all of this is our first surface area S1 that we found. So S2 is equal to 2S1. Therefore, the area of the surface generated when the graph of G on -33 is revolved above the x axis is twice the error generated when G on 03 is revolved above the X axis. A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Surface Area of Revolution

Symmetry of Functions and Intervals

Derivative and Its Role in Surface Area

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