Find the area of the region in the first quadrant bounded on the left by the y -axis, below by t

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

Key Concepts

Area Between Curves

This concept involves computing the region that lies between two or more curves by integrating the difference of functions. In calculus, when a region is bounded by curves, the area can be calculated by setting up an integral where one subtracts the lower function from the upper function, ensuring that the integration is performed over the correct interval. This method is a fundamental application of definite integrals in determining areas in the coordinate plane.

Determining Intersection Points

Before setting up an integral to find an area, it is crucial to determine where the bounding curves or lines intersect. These intersection points serve as the limits of integration, defining the range over which the area is calculated. Solving for these points involves equating the expressions for the curves and solving for the variable, which is essential to accurately describe the region in question.

Setting Up Integration Bounds

Once the intersection points are identified, the next step is to establish the appropriate bounds for integration. Depending on the region's geometry, one might choose to integrate with respect to x or y. Identifying the 'left/right' or 'top/bottom' boundaries through graphical analysis helps in determining the correct order of integration and ensures that the integral fully covers the region of interest.

Transcript

00:01 The area bounded by two curves on a closed interval from a to b is determined by, while the area a is equal to the integral going from a to b of, well, f of y minus g of y, dy, where we divide the shaded region into two parts, column a1 and a2.

00:25 And then we evaluate each of them separately and then add them up.

00:30 So for the limits of integration of a 1, we have 3 minus y is equal to 2 times the square root of y.

00:44 So we just square here, and this implies that 3 minus y squared is then equal to 4 y.

00:53 We just square both sides.

00:55 Get rid of the square root here.

00:56 Here.

00:58 And then, well, 3 minus y squared.

01:01 This is going to be, what, 9 plus y squared minus 6y, is equal to 4y.

01:06 We said this is equal to 0, and we have y squared minus 10y plus 9 is equal to 0.

01:18 This is going to factor as y minus 1 times y minus 9 is equal to 0.

01:26 Okay.

01:27 So now we have well, here we just take y is equal to 1 because y is equal to 9 doesn't satisfy here.

01:36 So we take y being equal to 1.

01:39 Okay.

01:40 And here we have a equal to 1 and b equal to 1.

01:44 And we have f of y is equal to 2 times the square root of y and g of y is equal to 0.

01:51 Therefore, we have a sub 1 is equal to the integral going from, 0 to 1 of 2 times the square root of y, d y, y.

02:04 Okay, so we evaluate this.

02:06 This is going to be equal to, well, we have 2 y to the 3 halves divided by 3 halves, evaluating from 0 to 1.

02:16 So we get 4 thirds, and then just times while 1 minus 0.

02:21 So this is equal to 4 thirds.

02:25 And then for the limits of integration of a sub 2, our second area here, well, we have y minus 1 squared is equal to 3 minus y.

02:42 So that implies that y squared, well, minus 2y plus 1 is equal to 3 minus y...

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