Doug C. answered • 08/17/18

Math Tutor with Reputation to make difficult concepts understandable

Claire H.

asked • 08/16/18Calculate the area of the region bounded by the two curves, y = 4 - x^{2} and y = x + 2.

The answer is meant to be 9/2 but I just cannot get it!!! Maybe my bounds are incorrect.... help!

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Doug C. answered • 08/17/18

Math Tutor with Reputation to make difficult concepts understandable

Hi Claire,

Take a look at this graph to get the idea of what is happening. This problem is an example of the "area between two curves". To find the height of a "typical rectangle" use "top" - "bottom".

https://www.desmos.com/calculator/mmynniltq1

Hopefully the above graph will give you some ideas on how to use Desmos to get a feel for what is happening for problems like this.

Mark M. answered • 08/16/18

Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.

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Claire H.

Sorry but I don't understand how you can just use 4-x^{2} and x+2 straight off the bat. Would you not have to split them because of where they "enter" and "exit" if you understand my meaning?

What I did, but it gives me the wrong answer is:

∫_{(fr}_{om 0 to 4) }_{∫ (from (y-2) to (-√(4-y)) dxdy + ∫(from 2 to 4) }_{∫ (from (y-2) to (√4-y) }

but this keeps giving me the incorrect answer and I cannot see why

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08/17/18

Paul M. answered • 08/16/18

Learn "how to" do the math and why the "how to" works!

I think that you mean the area bound ONLY by the parabola an the straight line.

Part of this area will have a negative sign; therefore you have to find the area in pieces.

The area enclosed by the parabola and the x-axis is twice the area bounded by the parabola, the y-axis and the x-axis to the right of the origin, i.e. twice the integral from 0 to 2 of 4 - x^{2}; that area is 32/3

The triangular area bounded by the straight line and x = 1 is 9/2.

the area bounded by the parabola, x = 1 and the x-axis is the integral of 4 - x^{2} between 1 and 2; this area is 5/3.

The area required then is (32/3) - (9/2) - (5/3) = 9/2.

Paul M.

tutor

First of all, you should make a graph of the area in question so that you can see what you are trying to compute.

Secondly, the reason you are having trouble is that the area in the 2nd quadrant will have a negative sign (because the abscissa is negative and the ordinate is positive).

You can solve the problem the way I did or you can integrate in 2 pieces.

You need to integrate the function which is the difference between the parabola and the straight line, between -2 and 0 and take the absolute value of the result and then add to that the integral between 0 and 1.

If what I have told you still doesn't help, post another "comment" and I will try to respond further.

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08/17/18

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Mark M.

^{2}= x+2. Solving this equation, we get x = -2 or 1.^{2}and the lower boundary is y = x+2._{(from -2 to 1)}[(4-x^{2}) - (x+2)]dx = ∫_{(from -2 to 1)}[2 - x - x^{2}]dx08/16/18