6 f(xy.=) dydzdx so that the orde Find the bounds for Question 6. [20 points] ` Include rough sketch Of the region. integration is dxdyd:...

$5-6$ Sketch the region whose area is given by the integral and
evaluate the integral.
$$
\int_{\pi / 2}^{\pi} \int_{0}^{2 \sin \theta} r d r d \theta
$$

The discussion will recall that the area under the girl can be written down in terms of the integral from A GP. Yeah, thanks. The X and in this question were given the integral from 0 to 6 six months X the X Yeah, we can identify This one will be the function f x. So let's try to scare the graph of the function. Yeah, and we have This will be in that excesses on we have Don't want access maybe. Yeah, x under why? And now we see when X equals zero Why could you six 36 help when x eco 261036 So therefore, we should have the crab. Looks like we need to change coupons together. So that's where the the function F X you go to the six months X here and the interval were given from 0 to 6 s. Well, so the hungry Shin Dong will are you looking for will be this one. I look at this area here within the next agenda, Changle, with the place equal to six and the high here, you could do six as well. So therefore we have the area. If we go to in the behalf, but it's times the height. So we get equal to you three times six they go to the 18.

So here on this problem we have to sketch the region whose area is given by the integral are they are digital. And the limit is from 1 to 2. And what three days by over four. 2. 3 pi over four. No, first it will be sketching though area. Now here we see that this, you know these limits are for our this was our equals to one, two articles to to No this uh these these limits are for theater is a theatre ranges from three d equals two pi over 4-3 d equals two through by awful. So by using these conditions we can easily sketch the region. So this shaded region is our the region which is defined by this integration. Now this is r equals to two And this red is r equals to one And here three d equals two by by four. And this Tedo this one this two days three by over four. So I have sketched the region. We had also evaluate this in trouble. So let me evaluate this. So we get the double integration. No the integration of our with respect to D. R. Would be our square over to So our square divide by two. And the limit is from 1 to 2. Got data. And the limit for ah the street up we have by over four 2, 3 pi over four. You can put the limits and after putting the limits upper limit minus or low limit would will get three over to data. And the integration By over 4- three by over four. No, this is constant. Three orders constant. So concerned with respect to theta will integrate so we'll get 3/2, 3 to and the limit will be from pi over 4 to 3 pi over four. Now we can put the limits three by two as it is. Upper limit is three pi over four. Mine is a low limited spy, or for so you can simplify this and finally, we'll get the result to be three Pi over four. Therefore, the The result for the given interval is three pi over four. So disregard answer. I hope you have understood the problem. Thank you.

So we're giving several equations. Y equals X. It's just a parent function, like was to oops so horizontal line. Then they give you why equal six minus X. So 3456 and then minus X Oh, I'm not graphing this very well. Why equals six minus X And what would be important to find out is these points of intersection Um oh, and they also give you y co zero. So I guess we don't even need the points of intersection. So we're looking at this area first. I'll sing. You were looking at that area, but that actually makes the problem really easy. Um, because if you do everything in terms of why Michael zero to y equals to the upper function is clearly this one Further to the right, I guess I should say, But you want to solve for X, So solving this for X will be adding X to the left side. Andan subtract wide to the right. So looking at six minus y and then subtract off the lower function, which is this Why equals X? So you just write minus y do I now might help you to write it as six minus two y do I. But you don't have to, because now you can find the anti drove over. This just six y minus. You add one to your exponents. Divide by two will cancel that out from 0 to 2. Well, you can plug in to then six times two is 12 two squared is four and you could do these minus zeros. But subtracting zero is not going to do anything. You just have 12 minus four, which equals eight, which if you took the trapezoid that they give you in the original problem, you start finding all these, break it up and find each area that will be really evident that the area is eight.

So this problem likes us to express the area of the shaded region as the sum of two inter girls with respect away. So when integrating with respect to y, sometimes it's helpful to envision the, um the image as tilted 90 degrees. So we also need to convert each of these formulas to be with respect, toe Why so right now, why equals two minus X? This is an equation with respect to X. After performing some simple algebra by solving for X, we can express X as a function of wine. Um, so this gives us quickly to minus y times tu minus y. And then why equaling X can quickly just be flipped around as X equals while. So let's go ahead and label these. Um, so where this guy right here is him and this is the new form of this red line Re so we want to integrate this shaded region whom we going highlight that for us, this shaded region here it's a little hard to see, But this green triangle we're gonna go ahead and integrate that with respect, toe. Why? Um and so this gives is kind of a funny shape. Now, But what we can do is we can split this down the middle, and now we have two triangles to integrate. Um, so this first triangle, let's call in our black Triangle because its corresponding to this black, um, this black function here. So that seems to be from to tow one x values. Why values? I mean to say sorry. So from 1 to 2, we're finding the area underneath the function tu minus. Why do you why And then for this other red triangle over here. We're integrating that from why Values zero toe one and the area under the curve. We're trying to find us the curve X equals y. So to find the area of this entire space, we simply take the sum of both of these integral 01 Why do you why? Plus wanted to do you wanna? So why do you and notice the problem is asking to asking for us to express this area off the shaded region as the sum of two inter girls which we've just found on. And there's no need to actually evaluate the integral. They just wanted us to set up a problem