[Technology Exercise] When solving Exercises 33-40, you may ne...

Question

[Technology Exercise] When solving Exercises 33-40, you may need to use a calculator or a computer.

Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.

y = x²/4, 0 ≤ x ≤ 2

Accepted Answer

Hello. In this video, we are going to consider the curve Y is equal to X2 divided by 16 that runs from X is equal to 0 to X is equal to 4. When this curve is revolved about the x axis, it generates a surface. We want to find the area of the surface generated by using the trapezoidal rule with three subintervals, and we want to round our answer to two decimal places. So, the first thing we want to do is we want to go ahead and generate or find the area of the surface. Now, the area of the surface is defined as the following. It is S is equal to 2 pi multiplied by a definite integral from A to B of Y multiplied by the square root of 1 + Y 2 DY. The first thing we want to do is we want to define our boundaries. Now we are told that the curve runs from X equal to 0 to X is equal to 4. So A is the smaller boundary, which is 0, and B is the larger boundary, which is 4. And we are told to use the curve y is equal to x 2 divided by 16. So we know that Y is defined as X2 divided by 16. And if we were to take the derivative of this function, we get Y is equal to x divided by 8. So now that we have the necessary parts we need to define the surface area, let's go ahead and define the surface area by using the information. So we get S is equal to 2 pi multiplied by the integral from 0 to 4 of X 2 divided by 16, multiplied by the square root of 1 plus X divided by 82, which is X2 divided by 64. D X OK, so now what we want to do is we want to go ahead and use the trapezoidal rule to estimate the value of the integral. Now the trapezoidal rule is defined as the following. It is the width of the subintervals divided by 2 multiplied by g of x0 plus 2 multiplied by g x1. Plus the infinite summation all the way to 2 multiplied by g of xn minus 1. Plus g of xn So, what we need to do is we need to go ahead and first calculate the widths of the sub-intervals. Now in this case here, delta X is defined as b minus a divided by n. It is the distance of our boundaries divided by the amount of subintervals. And keep in mind that we are working with 3 subintervals, so n is going to equal to 3. That means delta X will be defined as 4 minus 0 divided by 3, which is going to be 4/3. Now the next thing we want to do is you want to go ahead and define the values or the. End points of our interval from 0 to 4. So, we're going to start with the smallest endpoint, which is going to be X0. $$X_0 is $$going to equal to 0 for our starting value. Then we want to go ahead and add 4/3 to this value until we hit a total value of 4. So X1 is going to equal to 4/3. Adding another with the 4/3 gives us X2, which is going to be 8/3, and adding another value of 4/3 gives us X3, which is 16 divided by 3. Note 12 divided by 3, which goes ahead and gives us a value of 4. So these are going to be the endpoints of the function that we are working with. Now G of X in this case is going to be defined as the function of the integrand, that is going to be x 2 divided by 16 multiplied by the square root of 1 + x 2 divided by 64. So if we were to plug in the first point into the function, G 0 will have the value of 0. Plugging in the next end point gives us g of 4/3, and that is going to have a value of square root. 37 divided by 54. G of 8/3 gives us a value of 4 square root of 10 divided by 37. And G of 4 gives us a value of square root 5 divided by 2. So now if we were to take these values and plug them into the trapezoidal rule, the trapezoidal rule would be defined as the following. We have t is equal to 4/3 divided by 2. Multiplied by g of 0, which is 0, + 2 multiplied by g of 8/3. Which is 4. Or g of 4/3, which is going to be square root 37. Divided by 54. Plus 2 multiplied by g of 8/3, which is 4 square root of 10 divided by 37, plus g of 4, which is square root 5 divided by 2. Now, if we were to use a calculator, this is going to approximate to have a value of 1.520195. Now this is going to be the approximate value of the integral in the surface area equation, but keep in mind that the surface area equation is defined as 2 pi multiplied by the approximate value of the integral. So what this means is that now we're going to take this approximated decimal and plug it back into our surface area equation. That is going to give us s is equal. 2 2 pi multiplied by 1.520195, and by using a calculator, this is going to approximate to be 9.55, which is going to be the solution to the overall problem. And the overall solution is going to be option D. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

[Technology Exercise] When solving Exercises 33-40, you may need to use a calculator or a computer.Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.y = x²/4, 0 ≤ x ≤ 2

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[Technology Exercise] When solving Exercises 33-40, you may need to use a calculator or a computer.
Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.
y = x²/4, 0 ≤ x ≤ 2