Find the area of the surface generated when the given curve is...

Question

Find the area of the surface generated when the given curve is revolved about the given axis.

y=3x+4, for 0≤x≤6; about the x-axis

Accepted Answer

Welcome back everyone. In this problem, we want to determine the surface area formed when the curve Y equals 4 X + 1 for X greater than or equal to 2, but less than or equal to 5 is revolved above the X axis. Now what do we know about the surface area that's formed when a curve is revolved above an axis? Well, recall That the surface areas is going to be equal to 2 pi multiplied by the definite integral between the limits of A and B. of Y sorry, multiplied by the square root. Of 1 plus the derivative of y with respect to X all squared, all with respect to x. So in other words, if we can substitute our values here for A, B, Y, and the derivative of Y, then we should be able to plug it into the formula for the surface area and then solve, OK? So let's go ahead and do that. So far we know that our lower limit is 2, so A is 2, or upper limit B is 5. We were told that Y equals 4 X + 1, and we know that if we differentiate Y with respect to X, it will be equal. 4, OK. So in that case, if we plug all of these into our formula for the surface area, that means it will be equal to 2 pi multiplied by the integral between the limits of 2 and 5 of 4 X plus 1, OK, multiplied by the square root of 1 + 4 squad, OK, all with respect to X. If we go ahead and simplify this. OK. Then we're gonna have 2 pi multiplied by this or constant the square root of 1 + 4 squared. That's 1 + 16, which is 17. So we'll have 2 pi roots 17 multiplied by the integral between 2 and 5 of 4 X plus 1. OK, with respect to X. So now let's go ahead and integrate for X + 1 with respect to X and then see if we can solve. So now integrating. OK, so integrating. A term here In 2 and 5 of 4 X + 1 with respect to X we can integrate them separately. OK, so the integral of 4 X + 1 is 2 X squared and the integral of 1 is X. So we'll have 2 X2 + X between the bounds of 2 and 5. We know to solve that, we can substitute our bounds or our limits. So that will be 2 multiplied by 5 squared plus 5, OK. And all of that will subtract 2 multiplied by 2 squared plus 2. No, 25 squared is 25, 2 multiplied by 25 is 50, and 50 + 5 equals 55. On the other hand, 2 squared is 42 multiplied by 4 is 8, and 8 + 2 is 10. So this integral will be equal to 55 minus 10, which equals 45. Know that we have the integral, we can put it back into our formula for the surface area. So that means it will be equal to 2 pi route 17. Multiplied by 45, which is going to be equal to 905 multiplied by route 17. Thus The area surface area formed when the curve is revolved above the x-axis will be 90 route 17 square units. Thanks a lot for watching everyone. I hope this video helped.

Find the area of the surface generated when the given curve is revolved about the given axis.y=3x+4, for 0≤x≤6; about the x-axis

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Find the area of the surface generated when the given curve is revolved about the given axis.

y=3x+4, for 0≤x≤6; about the x-axis