Last HW problem looks so simple ... but cannot solve

Dec 2013
Find area enclosed by given graphs

y = 4 + root x
y = (12+x)/3

so, here we have the eq's solved for x

also, I solved for y and got x = (y-4)^2 and x = 3y-12

so looking at the graph I figured the rectangles would do better using f(y)

Thus, we have integration limits y^2 -8y +16 -3y +12 which simplifies to

y^2 - 11y + 28 which simplifies to

(y-7)(y-4) giving us a = 4, b = 7 or

if using c and d for variables

c = 4, d = 7

Now, INT from 4 to 7 of y^2 -11y + 28

= [(7^3)/3 - 11(7^2)/2 +28(7)] - [(4^3)/3 - 11(4^2)/2 +28(4)

Now to make sure I have not flipped signs incorrectly we have:

The first part minus the second part which looks like the first part

- (4^3)/3 + 11(4^2)/2 - 28(4)

So when I did this and got common denominators and combined like terms,

I ended up with a negative value for my final answer and I know if nothing else from this forum that you all have taught me that we cannot have a negative value for Area or Volume. Hence, I have done wrong or performed an error in my operations. So, I am asking for some help from the forum.


Forum Staff
Oct 2008
London, Ontario, Canada - The Forest City
\(\displaystyle 4+\sqrt{x}=\frac{12+x}{3}\)

\(\displaystyle 4+\sqrt{x}=4+\frac{x}{3}\)

\(\displaystyle 3\sqrt{x}=x\)

\(\displaystyle 9x=x^2\)

\(\displaystyle x(x-9)=0\)

so the bounds of integration are 0 to 9.

Now evaluate

\(\displaystyle \int_0^9x^{\frac12}-\frac{x}{3}\,dx\)

Answer: 4.5.


Math Team
Dec 2013
You got a negative answer because the curve is closer to the y-axis than the straight line, so your equation ought to be $-\left(y^2 - 11y + 28\right)$.

Otherwise, you seem to have got it. Well done!
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