Calculate the area between the two curves The two equations are $x=9y^2$ and $x=7+2y^2$. I made them equal to each other, found the two points which was $1,-1$. Then I calculated the integral subtracting both equations and got $-28/3$. This is wrong. Please help. Thanks
 A: Consider the graphs of the two functions:

From this we can see that for $-1\leq y\leq 1, 2y^{2} + 7 \geq 9y^{2}$
Therefore the area between the curves, $A$ is given by:
$A = \int_{-1}^{1}2y^{2}+7-9y^{2}dy = \int_{-1}^{1}7-7y^{2}dy$ 
$A= [7y - \frac{7y^{3}}{3}]_{-1}^{1}$
$A= 2(7-\frac{7}{3}) = \frac{28}{3}$
You got a negative answer because you didn't consider which function was larger for the domain you were integrating over.
A: The bounded region that we are looking is given in the shaded area below.

Now, in set notation, the region $R$ is given by $$R=\{(x,y):9y^2\leq x\leq 7+2y^2\text{ and } -1\leq y\leq 1\}.$$ Because of symmetry, we get
$$A=2\int_{0}^1\big[(7+2y^2)-9y^2\big]dy=2\int_{0}^1\big[7-7y^2\big]dy=2\bigg[7y-\frac{7y^3}{3}\bigg]_0^1=2\bigg(7-\frac{7}{3}\bigg)=\frac{28}{3}.$$
A: One thing to remember is that the area of a curve is always a positive number, since you subtract the bigger function from the smaller function, when integrating with respect to $x$, and rightmost function minus leftmost function when integrating with respect to $y$.
What you have done is got the right answer, but negative. This means that you have not subtracted properly.
Since we are integrating $w.r.t$ y, it's rightmost - leftmost.
Rightmost = $7+2y^2$
Leftmost = $9y^2$
Do rightmost - leftmost, and you will get $\frac{28}{3}$
