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Problem 19

Evaluate $\int_{C} x d s,$ where $C$ is a. the s…

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Problem 18

Integrate $f(x, y, z)=-\sqrt{x^{2}+z^{2}}$ over the circle
$$\mathbf{r}(t)=(a \cos t) \mathbf{j}+(a \sin t) \mathbf{k}, \quad 0 \leq t \leq 2 \pi$$

Answer

$$-4 a^{2}$$



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Video Transcript

in this problem were given a function F in the space curves see with parameter ization given by our and were asked to find the integral of f oversee. I first noticed that X squared plus C squared is always greater than or equal to zero. And therefore, the square root of X squared plus C squared is defined for all values in our three. So f is defined in our three and therefore f is defined on C. We're gonna break this up into two cases. So case one is if the parameter A equals zero in this case are is simply zero vector for all t. Therefore, the integral of Beth over a curve c is simply the integral of f over a point in our three the origin which is equal to zero because a 0 to 0 this is equal to negative four a. Squared her case to let's assume that a is not equal to zero. I noticed that f is a composition of continuous functions and therefore f is continuous. So f is continuous on. See notice also that both components of our our continuous and therefore are is continuous our prime simply the derivative of all the components of our. So this is equal to negative a sign T j hat plus a co sign t que hat and we see that our prime of tea is continuous. For all the tease in the domain of our this is because its components are both continuous For all T, Lastly, noticed that the norm of our prime of t squared This is going to be equal to the sum of the squares of the components. So we have negative a sign t squared, plus a co sign t squared. And then, using identities, you find that this is equal to a squared. Therefore it's not equal to zero. Thus, or a prime of tea is never zero. For if it were, surely it's normally be equal to zero. And so the square of its normal. So we know that our is continuous and our prime is continuous and our prime is never zero. So you can conclude that the curve c parameter as by our yes, smooth. Since F is continuous on C and C a smooth, we now know that the integral of f oversee exists and as a matter of fact, is given by this formula, we're taking the integral over the bounds of the parameter for our so from 0 to 2 pi and we're going to substitute for X, y and Z the life Jake in case components of our so zero a co sign t a sigh Inti times the Normie of our prime of tea DT Now this is equal to integral from 0 to 2 pi of the negative square root And then we know that x zero So x squared plus c squared is just C squared, which is a signed T squared times the norm of our prime of T. This is the square root of the norm of our kind of tea squared. So this is squared over a squared or the absolute value day U t now pulling the a out from under the radical we get, This is equal to negative absolute value. They squared integral from 0 to 2 pi of the absolute value of signed t DT. Now we know simple formula for the anti derivative sign of tea, but not really for the absolute value of sine of t. So let's use the additive ity of land girls to write. This is a sum of two integral. So this is equal to negative. A squared, since absolutely of a squared is the same as a squared integral from zero to pi of the absolute value of 70. Now, from zero to pi sign is always positive. So absolute value is the function itself d t plus the integral from pie two pi and then from pi to pie sign is always negative. So absolute value is the opposite of the function DT now taking anti derivatives You see that this is equal to negative a squared times negative coast sine of t and they redid it. Zero and pie plus co sign of tea. They needed a pie and two pi plugging in values. We get negative a squared times the opposite of co sign up eyes negative one minus coast Senate zero is one plus co sign of two pi is one minus Roseanna pies, Negative one. And simplifying this equals negative four Ace, where which is our answer. And you see this answer is independent of the value of a

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