Sketch the region enclosed by the curves and find its area

[destinynj]

Sketch the region enclosed by the curves and find its area

1. x=(2y)^2, y=(-1/2)x+2

2. y=x^2, y=2x+3

3. y=sinx, y=cosx, x=0, x=(3pi)/4

It would be greatly appreciated if you guys would help me understand how these are done instead of just giving the answers. Thank you so much!

 

[Deleted member 4993]

1. x=(2y)^2, y=(-1/2)x+2

2. y=x^2, y=2x+3

3. y=sinx, y=cosx, x=0, x=(3pi)/4

It would be greatly appreciated if you guys would help me understand how these are done instead of just giving the answers. Thank you so much!

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[stapel]

Sketch the region enclosed by the curves and find its area

1. x=(2y)^2, y=(-1/2)x+2

The first equation is a sideways parabola; the second is a straight line. Graph them the same way you learned back in algebra. (You may find it helpful to convert the first equation into two by solving for "y=", which will generate a plus-minus square root.)

To find the endpoints of the areas, solve the system of non-linear equations. (here) I would start with:

. . . . .\(\displaystyle y\, =\, \left(-\dfrac{1}{2}\right)\, (2y)^2\, +\, 2\)

. . . . .\(\displaystyle y\, =\, -2y^2\, +\, 2\)

. . . . .\(\displaystyle 2y^2\, +\, y\, -\, 2\, =\, 0\)

...and then apply the Quadratic Formula to find the y-values of the system's solutions. Then back-solve for the corresponding x-values.

Then using this information to set up the integral.

2. y=x^2, y=2x+3

This one works the same way as the previous one, though the set-up process will probably be simpler.

3. y=sinx, y=cosx, x=0, x=(3pi)/4

Here, they've given you the endpoints. But do the graph, too; I suspect you'll discover another "endpoint" where the first two curves cross.