A surface is generated by revolving the line f(x)=2−x, for 0≤x...

Question

A surface is generated by revolving the line f(x)=2−x, for 0≤x≤2, about the x-axis. Find the area of the resulting surface in the following ways.

a. Using calculus

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the surface area formed when the line F of X equals 5 minus X for X greater than or equal to 1, but less than or equal to 3 is revolved about the X axis. Now what do we know about lines revealed about the X-axis? Well, recall that the surface area is. Is going to be equal to 2 pi multiplied by the integral between the limits of A and B of F of X multiplied by the square root of 1 plus the derivative of F with respect to X squared all with respect to x. Now what from our problem can help us to find this information? Well, so far we know that a or lower limit is 1, B, the upper limit is 3. F of X equals 5 minus X, which means that the derivative of F with respect to X will be equal to -1. So if we plug these into our formula. Then that tells us the surface areas will be equal to 2 pie. Multiplied by the integral between 1 and 3 of F of X 5 minus X, OK. Multiplied by the square root of 1 plus the derivative of F with respect to X-1 squared or with respect to X. Now if we do some simplifying here, -1 squared is 1 and 1 + 1 is 2, so we have root 2, a constant inside our integral, and when we pull that constant out, we end up with 2 pi root 2 multiplied by the definite integral between 1 and 3 of 5 minus X with respect to X. Now let's evaluate the integral. So that means we're going to be integrating a 5 minus X with respect to X. Now we can integrate both of those separately. The integral of 5 is 5 X and the integral of negative X is negative X2 divided by 2. So that's 5 X minus 1/2 of X2d between the limits of 3 and 1. When we apply our limits, that's going to be 5 multiplied by 3 minus 3 square divided by 2. Minus 5 multiplied by 1 minus 1 squared divided by 2. Let's simplify this even further. In our first term, we're gonna have 15 minus 4.5, which is 10.5. And in the second, we'll have 5 minus 0.5, which is 4.5. OK? So that's 10 minus 4.5, OK? 10.5 minus 4.5, sorry, which equals 6. So a definite integral is 6. Therefore, when we substitute it or put it back into our surface era formula, the surface area will be 2 pi root 2 multiplied by 6, which equals 12 pi route 2. Therefore, the surface area that's generated by the curve revolved around the x axis or above the x-axis would be 12 per root to square units. Thanks a lot for watching everyone. I hope this video helped.

A surface is generated by revolving the line f(x)=2−x, for 0≤x≤2, about the x-axis. Find the area of the resulting surface in the following ways.a. Using calculus

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A surface is generated by revolving the line f(x)=2−x, for 0≤x≤2, about the x-axis. Find the area of the resulting surface in the following ways.

a. Using calculus