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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

Sagot :

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.

View image ujgerman
lhmarianateixeira

In this exercise we have to use the function informed to identify the value of the formed figure, then:

[tex]x=4-12y^2[/tex]

Given the following information in the exercise statement:

  • [tex]2x+y^2=8[/tex]
  • [tex]y=x[/tex]

Substituting y for x in the equation:

[tex]2y+y^2\\y^2+2y-8(y+4)(y-2)=8[/tex]

Analyzing the equation formed, we can see that it will have the shape of a parabola, and that the values ​​of x will be between:

[tex]X: -4\leq y\leq 2[/tex]

See more about functions at brainly.com/question/5245372

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