Find the area of the region in the first quadrant bounded on...

Find the area of the region in the first quadrant bounded on the left by the $y$ -axis, below by the curve $x=2 \sqrt{y},$ above left by the curve $x=(y-1)^{2},$ and above right by the line $x=3-y$
GRAPH CANT COPY

Key Concepts

Graphical Analysis

Sketching a rough diagram of the curves and the bounded region aids in visualizing the problem. This analysis not only clarifies the relationships between the curves but also assists in correctly identifying the integration limits and the overall structure of the solution.

Region of Integration

This concept involves identifying the area bounded by several curves in the coordinate plane. In multivariable calculus, it is important to understand how a region is defined by different functions so that the correct area or volume can be computed through integration.

Intersection of Curves

Finding the points where the bounding curves intersect is crucial because these intersection points determine the limits of integration. Solving the equations of the curves gives the precise coordinates at which the boundaries meet, which are then used to set up the integral properly.

Setting Up the Definite Integral

Once the region and its boundaries are clearly understood, the next step is to express the area as a definite integral. This involves choosing the correct variable of integration and establishing the integration limits based on the geometry of the region.

Order of Integration

Deciding whether to integrate with respect to x or y first can simplify the problem significantly. The bounds of the region, as defined by the given curves, help determine the more natural variable for integration, thereby making the integral easier to evaluate.

Transcript

00:01 We want to find the area with that graph that they give us shaded in.

00:05 So let's look at the region that we have.

00:09 So everything is in terms of y, so it's probably a good idea to just integrate with respect to y.

00:17 And notice that we can break this up actually into two regions right here where y is equal to 1.

00:26 And it would be, so we can call this first region up here 1.

00:31 And i'll call the second region 2.

00:33 So the first region would be, so we start on the left here, and we go and enter on the red, and then we pass through the blue.

00:45 And from the graph they give us, it's pretty easy to see that y ranges from 1 to 2.

00:51 The blue line is going to be our upper bound.

00:54 So it's going to be 3 minus y, and the red is going to be our lower bound, so minus 1 squared, dy.

01:02 And actually probably need to expand the screen a little bit.

01:09 And then for the next area, region 2, so we start over on the left, so we pass through at x is equal to 0, and then we pass through the green.

01:24 So that looks like y ranges from 0 to 1 in that region, and we would have 2 root y minus 0, which wouldn't do anything, so we could just leave that off.

01:37 So we would integrate these two right here to get what our area for that region is.

01:45 So let's first go ahead and expand what we have there.

01:52 So it would be the integral from 1 to 2 of 3 minus y.

01:58 So then that should give us negative y squared...

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