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 NOT 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.

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.

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.

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.

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.

Transcript

00:01 So since they give us everything in terms of y or sorry x equals, it actually makes sense to, you know, do things in terms of y.

00:10 So the first integral you can see in terms of y goes from zero to one.

00:15 And that's just the function two square root of y.

00:19 I would write as y to the one half power.

00:22 And then when you do the integral from one to two, because everything's in terms of y, the upper function is, or furthest to the right, i guess i should say is the 3 minus y function and then subtract off the other function that's to the left which is that y minus 1 being squared d y now before doing this problem i would clean up this function just because i don't really like that y minus 1 being squared so rewrite and you foil it out be y squared minus 2y plus 1 but don't forget that you you're subtracting off each one of those terms.

01:04 So as i distribute that in, it's going to become negative y squared and negative y plus 2y and then 3 minus 1 plus 2, d .y.

01:22 And then it's pretty easy now, i think, just to do each problem, each piece.

01:29 2 y to 1 half, dy.

01:31 And then add the areas together.

01:33 So as you add one to your exponent, three halves and divide by your new exponent or multiply by the reciprocal two times two is four thirds, and that's from zero to one.

01:47 And then add to it this, which is negative one third, y cube plus one half, y squared plus two y, and that's from one to two.

02:01 So this is nice.

02:03 Just plug in one.

02:04 Any power is one.

02:05 So you just have four thirds there.

02:08 And then plugging in two, this is going to get kind of ugly for me...

Please give Ace some feedback