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Find the area bounded curve defined below on the Questlon #10 interval 0 <t <2tx(t) =-2cost y(t) = sin t...

Question

Find the area bounded curve defined below on the Questlon #10 interval 0 <t <2tx(t) =-2cost y(t) = sin t

Find the area bounded curve defined below on the Questlon #10 interval 0 <t <2t x(t) =-2cost y(t) = sin t



Answers

Find the area of the region bounded by hypocycloid $\mathbf{r}(t)=\cos ^{3}(t) \mathbf{i}+\sin ^{3}(t) \mathbf{j} .$ The curve is parameterized by $t \in[0,2 \pi].$

Here. We're looking at the area underneath the curve of signed two X going from zero to pi over four. Notice that on the center of all the function signed, two X is always good than or equal to zero. Now let's go ahead and integrate. Get negative. One half co sign to t evaluating a power 40 Get negative one half zero minus one. So that's just one half.

But I'm a tricky questions. He'll he's good people. Good. Okay. From zero of heaven toe spdr mhm to the mhm. Mhm. Yeah. Fucking less clear with people before mhm with the the local community by your peers cause Scurti, Of course. De Okay, this is the current tickets. Zero toe pregnant toe because cure activity. Yeah, so, upon performing integration of this yet, sign de a minus. Thank you. T v three. This is a photo ableto this is a global signed a very to heartless thank you 5 to 3 madness. Say this Eagle plus sign. Seattle three. This is a proto one minus by three. This is a photo by three. The radio as you go to buy three.

We wish to find the area under the curve. Why is equal to one over coastline square T. And this is between he is equal to zero and T. Is equal to pi over two. So to find the area under the curve, we need to go ahead and find the interrupt. So what we're looking for is the integral from zero to pi over two of our curve. So one over coastline square T. Can be written as second squared T. So now this is an improper integral because when we try to put in pi over two we get um 1/0, right? Because co sign of high over two is zero. So now this is one divided by zero. So that means the first thing we need to do is we need to replace that with a dummy variable. So we're going to say the limit as a approaches. Hi over two of the integral from zero to a. Of seeking square T. And now we can go ahead and find the integral. So the anti derivative of second squared is tangent. So therefore this is equal to the limit as a approaches high over to of tangent. Okay, evaluated from zero to A. So this is equal to limit. As a approaches pi over to put in um substitute in our limits of integration. And you should get this is tangent but a minus tangent of zero. So 10 0. Well, that's just zero and tangent of A as a approaches high over two is going towards infinity. So that means this is divergent. So what this means is the area under this curve between zero and pi over two is divergent. So it doesn't have a limit, it's going towards infinity.

I want to find the area under the curve C. Of T. On the interval zero pie or two. So we'll use this equation nine formula up here. So area is the integral from 0 to Pi over two Y. Of T. Which is the cosine squared of T. Times dX DT and the derivative of the sine of T. Is the coastline of T. D. T. All right. I'm going to take out the coastline squared and put in one minus sine squared so that I can use this coastline. T. S. D. U. Zero to pi over two. One minus sine squared T. Co sign T. D. T. Which is really too in a girl zero to pi over 21 times the coastline T. So co sign T. D. T zero Pirate too. Sign square T. Go sign T. D. T. All right. The integral with the coastline of T. It. It well is the sign of T. From 0 to Pi over two here. I am letting you be the sign of tea than do you? Is the coastline of T. D. T. So this is really you squared do you? So it's in its early U. Cubed over three. So minus the sine cubed of T. Over three. From 0 to Pi over two. Okay so this is the sign of Pyro or two. When is the sign of 0? Sign of power to one minus sine of zero. Which is zero minus the sine of pi over two cubed over three minus the sine of zero cubed over three. Okay so we get 1 -1 3 two thirds.


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