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Use the arc length formula to find the length of the curve $ y = \sqrt{2 - x^2} $, $ 0 \le x \le 1 $. Check your answer by noting that the curve is part of a circle.

$\frac{\pi \sqrt{2}}{4}$

Applications of Integration

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we'll be looking at Why equals radical to minus X squared will want to use the arc length formula, which has a why prime in it. So we'll want to differentiate, what with respect to X. So we get negative X over radical to minus X squared. So we're on the interval. X is greater than or equal to zero, but less than or equal to one. So we will use that in arc length formula, putting in zero and one for our A and B one plus plugging in. Why prime here squared t X and then we can simplify this. Let's go to the next page. Um, we can see that X squared er negative X squared is going to be X squared over to minus X squared because squaring the radical cancels out and then plus one next will come over here and adding the one we can just say one. This is another way we can write one so that we can adhere. Um, thes cancel out and we're left with two over two minus X squared under the radical still D X equal. Come to the next page here so we can pull out the radical, too. And we're left with the integral 0 to 11 over radical to minus X squared DX. And we can see that this corresponds to a substitution we can make with Mark Sein. You over a So once we make the substitution, we can see that it comes down to radical, too. Sign our ark Sign X over radical too from 120 So we can then plug these in which I did on the next page, we can see that, um, we'll plug in one over radical too. And we can pull out the radical too first and then subtract arc sine one of a radical to minus arc science zero, which is just zero. So arc sine won over. Radical to is high over four minus euro. Not forgetting about that radical, too. On this equals pi ride too over four. And we'll go to the next page here to verify that this is actually an eighth of a circle. So the why equals rad to minus um X squared is the graph of the upper 70 circles under dhe at the origin with a radius of Brad, too. So, on this interval, um, dark length is 1/4 of the earth clink of the semi circle. So it is an eighth of the circumference of the whole circle. Um, which would be to pie are so we can verify here that's plugging in a radius of red two will just get us the answer we had before.