Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle.
2x + y^2 = 48, x = y

Answer 1

Answer:

Attached please find response.

Step-by-step explanation:

We wish to find the area between the curves 2x+y2=8 and y=x.

Substituting y for x in the equation 2x+y2=8 yields

2y+y2y2+2y−8(y+4)(y−2)=8=0=0

so the line y=x intersects the parabola 2x+y2=8 at the points (−4,−4) and (2,2). Solving the equation 2x+y2=8 for x yields

x=4−12y2

From sketching the graphs of the parabola and the line, we see that the x-values on the parabola are at least those on the line when −4≤y≤2.