Surface area limited by 2 functions: x = y^2 - 8 and x = -2y + 1

student127

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Hello, I need to find the surface area of a curved trapeze that is limited by the following functions; x = y^2 - 8 and -2y + 1. For an example like y=4X, y=0, x= 1 it is easy to do and would just be calculating integral of 4x and it's defined integral within bounds 0-1 and then using Newton-Leibinz formula. But in this example the functions are being defined against x and this confuses me. https://gyazo.com/6285709adfb42ddd9dfb2193c92dc901 is the kind of graphic it should be.Using symbolabs integral applications I got this: https://www.symbolab.com/solver/int...\left(y\right)=y^{2}-8, f\left(y\right)=-2y+1 But the graph isn't the same there, and also the bounds aren't the same: Bounds should be from -8 to 9.325 right?

How must I go about finding the surface area of this type?
What must I integrate and calculate the definite integral of?
Is the symbolab example correct and I just don't understand why?

Any help appreciated,
Best regards
 
The natural way to find this area is to integrate with respect to y rather than x. That is what the program is doing, though it seems to be sometimes saying so, and other times (including the graph) transposing the variables so that it is finding the area between y = x^2 - 8 and y = -2x + 1 (which will, of course, have the same area).

The limits of integration are values of y, not of x (in the original formulation), and they are correct.

Are you not familiar with using horizontal elements of area, integrating with respect to y? Here is an example: http://www.pleacher.com/mp/mlessons/calc2006/day109.html.
 
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