calculusintegration
Sketch the region enclosed by the curves. Decide whether to integrate with respect to x or y. Then find the area of the region between $$x=9-{y^2}$$ and $$x={y^2}-9$$
What I have done so far is put the equations in terms of y and make them equal to each other. I got $$9-x=x+9$$ which goes to $$2x=0$$. I am unsure of how to find the bounds of the integral from this. Can anyone help?
Here is the sketch.
You can count the squares and part squares to see the answer is a little over 20.
From Wolfram Alpha, we can sketch the curves to find the area of interest:
Note that we need to find the points of intersection: at $x = 0$ the lines $y = x, \;y = \frac x{4}$ intersect. At $x= 1$, the lines $y = x$ and $y = \frac 1x$ intersect. At $x = 2,$ the lines $y = \frac 1x $ and $y = \frac x4$ intersect. You can solve this by integrating between the relevant curves from $x = 0$ to $x = 1$, and likewise integrating between the relevant curves between $x = 1$ and $x = 2$, then summing: $$\int_0^1 \left(x - \frac x4\right)\,dx \quad + \quad \int_1^2 \left(\frac 1x - \frac x4\right)\,dx $$
Best Answer
You need to find the intersections of the two parabola. These occur when $9-y^2=y^2-9$, which is equivalent to $y^2-9=0$. So you have two intersections, namely $(0,-3)$ and $(0,3)$.
Draw a picture.
The area is given by $$ \int_{-3}^3(9-y^2-(y^2-9))dy=2\int_0^3(18-2y^2)dy. $$
I guess you can finish the calculation.