00:01
So for this problem, we are first asked to sketch the region and closed by the curves 2x plus y squared equals 48 and x equals y.
00:08
Then we are to decide whether to integrate with respect to x or y, draw a typical approximating rectangle, and then find the area of the region.
00:16
So we can see that in this situation, we tried integrating over x, then we'd have the complication of the upper bound essentially constantly changing here.
00:27
What we can do instead and have a much easier time is integrate first over, or not just first over, but integrate over y instead, where we would have that these are our sort of typical approximating rectangles.
00:50
So we would have that the area of the region is going to be equal to the integral from negative 8 up to 6 of, so actually, i'll note here, if we're integrating over y, we want x as a function of y.
01:08
So we can rearrange these equations.
01:10
We'd get that x equals 48 over 2, so 24, minus y squared over 2, and then we'd have x equals y...