Find the area of the region enclosed by the curves X + 2, Y squared equal 2 and X + y to the 4th equal 1 for X greater than or equal to 0. So to start this problem we need to figure out where these two curves are going to intersect. So I'm going to solve both of them for ** equal to -2 Y squared and X equal 1 -, y to the 4th, setting those two equal. In solving, I'm going to take everything to one side, Y to the 4th minus two, I ^2 + 1 = 0, factoring that y ^2 + 2 and y ^2 - 1. the Y ^2 + 2 is no real solutions. the Y ^2 - 1 gives us positive and -1. We can see that we're going to have symmetry here. So when I do the integral, I'm going to do 2 times the integral zero to 1. If you wonder about how we found the symmetry, think about if you put a negative Y for each of those, negative Y would end up coming out as the exact same equation as the original equation in both of them. When you simplify, the next thing I need to do is I need to figure out which equations further to the right or which one has the bigger X value for a specific location. So I'm going to look at both those equations and I'm going to put in a value somewhere between zero and one. I chose 1/2, so X equal 2 -, 2 to the 1/2 ^2 and X equal 1 - 1/2 to the 4th. I can see that the X equaling 3 halves is bigger than the X equaling 15 sixteenths, or three halves is further to the right than 15 sixteenths. So when I put this in, I'm going to have the 2 -, 2 Y squared equation coming first minus the 1 -, y to the 4th equation. We're doing this in terms of DY, so our Y values are going to go from zero to 1, and I'm going to multiply by two for the symmetry. I could have just as easily have gone from -1 to one. When we simplify that, we get Y to the 4th minus two I ^2 + 1 DY, we're going to take the integral and we're going to get two times the quantity 1/5 Y to the 5th -2 thirds y ^3 + y from zero to one stick in the upper bound minus the lower bound. We end up with 16 fifteenths.