8 4 pts evaluate each given integral by identifying it as an...

8. (4 pts) Evaluate each given integral by identifying it as an area or a volume. No credit will be given for calculator output or standard computation using the Fundamental Theorem. (a) $$ \int_{0}^{1} \int_{1-\sqrt{1-x^{2}}}^{2-x} d y d x= $$ Sketch or describe the area or volume this integral represents: (b) $$ \int_{0}^{1} \int_{1-x}^{\sqrt{1-x^{2}}} \int_{-1}^{5} d z d y d x= $$ Sketch or describe the area or volume this integral represents: 9. (3 pts) Evaluate the integral by reversing the order of integration. (Please show all steps in your evaluation. A substitution method may be used. No credit will be given for calculator output without the necessary work.) $$ \int_{0}^{6} \int_{x / 3}^{2} x \sqrt{y^{3}+1} d y d x $$

Transcript

00:01 All right, so for this problem, we are given an integral, double integral, and we have to change the order of integration.

00:10 So our first part, for a, we're given the integral from 3 to 5 of the integral of 0 to 2x minus 6 of f of xy, d, y, d, d, d x.

00:34 Now to change this, i want to draw a little graph and know what all of these things, the graph of all these things look like.

00:48 So, since this d .y first, these are our y equals.

00:53 So we have the line y equals zero.

00:56 I know what that is.

00:59 And the line y equals 2x minus 6, like that.

01:14 And now we have our x lines.

01:20 So x equals 3 is this line right here, and then x equals 6, or 5 rather, is right there.

01:31 So now we know that we're going to be looking at this area right here.

01:39 If we want to change it to be dx, d, y, then we have to go from horizontal.

01:54 You have to do horizontally what starts at on the left to what it goes to on the right, and then from top to bottom.

02:01 So what starts out on the left is this diagonal line, but we have to make it in x equals.

02:08 So just solving that for x, y plus 6 equals 2x, divide by 2, x equals 1⁄2, plus 3.

02:19 So that's the bottom range of the x.

02:26 And then the top range is 5.

02:33 Now we just have to see what goes down to on the vertical.

02:38 The lowest it goes is zero.

02:40 And the highest it goes is this point right here, which is at x equals 5.

02:46 But we're looking for the y value of it.

02:48 So just put a 5 in for x.

02:50 We should get 2 times 5, which is 10, minus 6, which is 4.

02:55 So there's our answer for problem 1.

02:58 For b, we have 0 to 1, and e to the v, and e, f of u v, d, d, u, v.

03:23 So now, the way we kind of thing about this is, so this is our x, this is our y, so really, this is x equals, it's u equals, but in our graph, we'll just have x equals.

03:41 So u equals e to the power of v looks like, let's change of colors, looks like this, sorry, that would be y equals, or v equals e to the u.

04:05 It looks like this, this is u equals e to the v, and then the u value, again remember this is u, this is at a, the u value of e, the u value of e.

04:18 Then we have our v values of 0 and 1, which is going to be right where those intersects.

04:37 So then here's our little area we're going to be dealing with.

04:42 Now we want dv first, so we want it to be parallel to that v axis, which is the vertical one.

04:59 So we want our going like this.

05:02 So the lowest we have is 0.

05:04 The highest we have is that curve, but we want to be v equals.

05:09 So u equals e to the v.

05:12 In order to get v, it would have to be v equals natural log of u.

05:20 Then farthest left, u value we have is zero...