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=4x−1, for 1≤x≤4; about the y-axis (Hint: Integrate with respect to y.)

Accepted Answer

Welcome back, everyone. In this problem, determine the surface era formed by revolving the curve Y E equals 5 X + 4, for X greater than or equal to 1, but less than or equal to 6 above the Y axis. Now how can we figure this surface area out? What do we know? Well, recall that the surface area generated by revolving a curve above the y axis S is equal to 2 pi multiplied by the definite integral between the limits of A and B of X. Multiplied by the square root of 1 plus the derivative of y with respect to x squared all with respect to x. Now do we know any of this information in our surface area formula? Well, we know that our lower limit A is one. Our upper limit B is 6. We know that Y equals 5 X + 4, so that means the derivative of Y with respect to X will be equal to 5. So we can use all of this information to solve for our surface area. Because no, that means it's going to be equal to or the surface here will be equal to 2 pi multiplied by the definite integral between 1 and a 6 or the square root of 1 + 5 squared, OK. Oh sorry, X multiplied by the square root of 1 + 5 squared, OK? All with respect to X. Now, we can simplify and solve. Now we know that 5 squared is 25, 25 plus 1 is 26, and route 26 is a constant. So we're going to have two pi route 26 multiplied by the integral between 1 and 6 of X with respect to X. Now we can integrate X with respect to X and the salve. So now when we integrate that, that's 2 by route 26 multiplied by X2d divided by 2, OK, between the boundaries of 1 and 6. And when we solve for it between those boundaries, OK, that's going to be equal to 6 squared divided by 2 minus 12 divided by 2. OK. 6 squared is 36, 1 squared is 1. So this would be 35 divided by 2. So that's 2 route 26 multiplied by 35 divided by 2. To Ks here. And no, this just leaves us with 35 i. OK. 35 i multiplied by the square root of 26 square units. So this, OK, is the surface area of our curve that's revolved around the y axis. 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=4x−1, for 1≤x≤4; about the y-axis (Hint: Integrate with respect to y.)

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

y=4x−1, for 1≤x≤4; about the y-axis (Hint: Integrate with respect to y.)