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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

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    Appart from a sign issue, why is the result you got wrong? – 2017-02-26
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    One crucial point in this kind of computation is to check (using a figure for example) which curve is above the other one. Did you do it? – 2017-02-26
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    It's good except area is always nonnegative, we need to integrate (7+2y^2)-9y^2 from -1 to 1. – 2017-02-26

3 Answers 3

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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 $\frac{28}{3}$

2

Consider the graphs of the two functions:

enter image description here

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.

2

The bounded region that we are looking is given in the shaded area below.

enter image description here

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

$$A=2\int_{0}^1\big[(7+2y^2)-9y^2\big]dy=2\int_{0}^1\big[7-7y^2\big]dy=2\bigg[7y-\frac{7y^3}{3}\bigg]_0^1=2\bigg(7-\frac{7}{3}\bigg)=\frac{28}{3}.$$