calculus
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to $x$ or $y$. Then find the area of the region bounded by $x+y^2= 6$ and $x+y=0$
this is what i got but it i wrong $6\cdot 3 – 3^3/3 – 3^2/2 – 6\cdot (-2) – (-2)^3/3 – (-2)^2/2$
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.
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
Here is the sketch.
You can count the squares and part squares to see the answer is a little over 20.