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We will find this area by integrating with respect to y. The integrand is obtained by taking the right-hand function minus the left-hand function, or (6 - 6y^2 - (6y^2 - 6)). Step 3 The limits on the integral are the y-values where the curves intersect. Equating 6 - 6y^2 = 6y^2 - 6, we find that the two solutions are y1 = -1 and y2 = 1. Step 4 Now, the area is given by ?_{-1}^{1} [(6 - 6y^2) - (6y^2 - 6)] dy = ?_{-1}^{1} (-12y^2 + 12) dy. Step 5 We have ?_{-1}^{1} (12 - 12y^2) dy = [72 - 72 x]_ {-1}^{1}. Performing this operation and simplifying fully gives us the exact area of the region, 0.

          We will find this area by integrating with respect to y.
The integrand is obtained by taking the right-hand function minus the left-hand function, or
(6 - 6y^2 - (6y^2 - 6)).
Step 3
The limits on the integral are the y-values where the curves intersect.
Equating 6 - 6y^2 = 6y^2 - 6, we find that the two solutions are y1 = -1 and y2 = 1.
Step 4
Now, the area is given by
?_{-1}^{1} [(6 - 6y^2) - (6y^2 - 6)] dy = ?_{-1}^{1} (-12y^2 + 12) dy.
Step 5
We have
?_{-1}^{1} (12 - 12y^2) dy = [72 - 72 x]_ {-1}^{1}.
Performing this operation and simplifying fully gives us the exact area of the region, 0.

We will find this area by integrating with respect to y.
The integrand is obtained by taking the right-hand function minus the left-hand function, or
(6 - 6y^2 - (6y^2 - 6)).
Step 3
The limits on the integral are the y-values where the curves intersect.
Equating 6 - 6y^2 = 6y^2 - 6, we find that the two solutions are y1 = -1 and y2 = 1.
Step 4
Now, the area is given by
?-1^1 [(6 - 6y^2) - (6y^2 - 6)] dy = ?-1^1 (-12y^2 + 12) dy.
Step 5
We have
?-1^1 (12 - 12y^2) dy = [72 - 72 x] -1^1.
Performing this operation and simplifying fully gives us the exact area of the region, 0.

Added by Tina H.

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