Consider the region bounded by the graphs of y = sin⁻¹(x), y =...

Question

Consider the region bounded by the graphs of y = sin⁻¹(x), y = 0, and x = 1/2.

b. Find the centroid of the region.

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find x bar of the region bounded by y is equal to inverse cosine of x. Y is equal to 0, x is equal to 1/2, and x is equal to 1. Use the table of intervals. OK, so it appears for this particular prompt we are asked to ultimately solve for x bar. So we are asked to find what x bar is of the region that is bounded by y is equal to inverse cosine of x, y is equal to 0, and when x is equal to 1/2 and x is equal to 1. And we're asked to accomplish solving for X bar by using the table of intervals. So that is our big hint. So now that we know that we're ultimately trying to solve for the value of X bar. Our first step that we need to take is we need to recall and note that the X coordinate of the centroid X bar for a planar region between Y is equal to F of X. And the x axis, which is Y is equal to 0, with a constant density on square brackets of A, B, will be given by the Fawley equation, which from our prior knowledge, we need to recall that X bar will be equal to MY divided by A. Fantastic. And we need to note that the moment about the y axis, which is denoted as MY, and the moment about the y axis MY will be equal to the interval evaluated from A to B of X multiplied by F of XDX. And A will represent the area, and the area A will be equal to the integral evaluated from A to B of F of X DX, and for this particular prom, we need to let Fx be equal to inverse cosine of x. So with that in mind, we now need to find the area, so we need to solve for area. So the area A will be equal to the integral evaluated from 1/2 to 1 of inverse cosine of XDX. And we need to recall a note that when we look at our table of integrals, we will notice that the integral of inverse cosine of AXDX will be equal to x multiplied by inverse cosine of AX minus 1 divided by A. Multiplied by the square root of 1 minus a 2 multiplied by x 2 plus C. Fantastic. So, Use using that piece of information using our table of intervals, then go ahead and write that the area A will be equal to the interval evaluated from 1/2 to 1 of inverse cosine of x dx. Will be equal to square bracket of square brackets of x multiplied by inverse cosine of x minus the square root of 1 minus x to the power of 2, and this expression will be evaluated from 1/2 to 1. And this will be equal to 1 multiplied by cosine, specifically 1 multiplied by inverse cosine of 1 minus the square root of 1 minus 12 minus square brackets of 1/2 multiplied by inverse cosine of 1/2 minus the square root of 1 minus 1/2 to the power of 2. And we will find when we plug this expression into our calculator and we write our final answer in its most simplified form, meaning we can't simplify it any further, we will find that it's simply equal to 3 multiplied by the square root of 3 minus pi divided by drum roll 6. Fantastic. Now, for our next step, we need to solve for the moment about the y axis. So we need to solve for MY. So My will be equal to the integral, evaluate from 1/2 to 1 of x multiplied by inverse cosine of XDX. And we need to recall a note from our table of integrals. We need to note that the integral of x to the power of n multiplied by inverse cosine of ax dx will be equal to x to the power of n + 1 divided by n + 1, multiplied by inverse cosine of ax plus a divided by n + 1. Multiplied by the integral of x to the power of n + 1, d x divided by the square root of 1 minus a 2 multiplied by x 2. Where we need to note very importantly that n cannot equal -1. Thus we can go ahead and write that my will be equal to the interval, evaluate from 1/2 to 1 of x multiplied by inverse cosine of x. The x, which will be equal to x to the power of 2 divided by 2, multiplied by inverse cosine of x, evaluated from 1/2 to 1. Plus 1/2 multiplied by the integral evaluated from 1/2 to 1 of x to the 2 dx divided by the square root of 1 minus x 2. So, from our table of intervals, we need to recall that the interval of X 2 DX divided by the square root of a 2 minus x 2. Will be equal to a 2 divided by 2 multiplied by inverse sin of x divided by a minus 1/2 multiplied by x multiplied by the square root of a 2 minus x 2. Plus c or c denotes some constant variable. So thus we can go ahead and write safely that my will be equal to the integral evaluated from 1/2 to 1 of x multiplied by inverse cosine of x d x. Which will be equal to square brackets of x 2 divided by 2 multiplied by inverse cosine of x plus 1/2 multiplied by square brackets of 1/2 multiplied by inverse sin of x minus 1/2 multiplied by x multiplied by the square root of 1 minus x to the power of 2. Evaluated from 1/2 to 1, which will be equal to square brackets of x 2 divided by 2 multiplied by inverse cosine of x plus 1/4 multiplied by inverse sin of x minus 1/4 multiplied by x multiplied by the square root of 1 minus x to the 2, and this will be evaluated from 1/22 to 1. So considering our at x equals 1 condition, that means we're plugging in a value of 1 in for x, we will have 12 divided by 2 multiplied by inverse cosine of 1 plus 1/4 multiplied by inverse sin of 1. And we will need to subtract 1/4 multiplied by 1, multiplied by the square root of 1 minus 1 to 2, which will be all simply equal to pi divided by 8. And for the condition x is equal to 1/2, we will find that 1/2 to the power of 2 divided by 2 multiplied by inverse cosine of 1/2 plus 1/4 multiplied by inverse sin of 1/2 minus 1/4 multiplied by 1/2 multiplied by the square root of 1 minus 1/2 to the power of 2. Will be all simply equal to negative, the square root of 3 divided by 16, when we plug that expression into our calculator. So therefore we could safely say that my will be equal to, so m subscript y will be equal to pi divided by 8 minus parentheses of pi divided by 12 minus the square root of 3 divided by 16. Which will be equal to when we write our final answer in its most simplified form, meaning we can't simplify any further, 2 pi plus 3 multiplied by the square root of 3 divided by 48. Thus, that will mean that x bar will be equal to my divided by a, which will be equal to 2 pi plus 3 multiplied by the square root of 3 divided by 48 divided by 3 multiplied by the square root of 3 minus pi divided by 6. Which will be all equal to 2 pi plus 3 multiplied by the square root of 3, all divided by 8 multiplied by parentheses of 3 multiplied by the square root of 3 minus pi. And that's it. That is our final answer. Hooray, we did it. So going back to look at the prom itself we could safely say without a doubt that our final answer for x bar is once again. 2 pi plus 3 multiplied by the square root of 3 divided by 8 multiplied by the parentheses of 3 multiplied by the square to 3 minus pi, and that's it. That is our correct and final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Consider the region bounded by the graphs of y = sin⁻¹(x), y = 0, and x = 1/2.b. Find the centroid of the region.

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Consider the region bounded by the graphs of y = sin⁻¹(x), y = 0, and x = 1/2.
b. Find the centroid of the region.