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I have two functions and I need to find the area between them: y^2=4x and x=3. I calculated the integral of 2*sqrt(x) on the interval [0,3] and I got 6.9282 for an answer. According to my homework solutions, this is incorrect. Is there something wrong in my setup or am I calculating something wrong?

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The region extends both above and below the $x$-axis; your calculation caught only the upper half. –  Brian M. Scott Dec 12 '12 at 5:04
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The upper boundary of the region is the curve $y=2\sqrt x$, but the lower boundary is $y=-2\sqrt x$, not $y=0$. You calculated the area of the upper half of the region.

This is a good example of a problem that can be done about as easily with $y$ as the independent variable. Instead of chopping the region up into ‘infinitely narrow’ vertical strips, chop it up into horizontal strips instead. The line $x=3$ intersects the parabola $y^2=4x$ at $y=\pm2\sqrt 3$, so your $y$-values will range from $-2\sqrt3$ to $2\sqrt3$. The horizontal strip at height $y$ runs from a low $x$-value of $\frac{y^2}4$ to a high $x$-value of $3$, so its length is $3-\frac{y^2}4$, and the area is given by the integral

$$\int_{-2\sqrt3}^{2\sqrt3}\left(3-\frac{y^2}4\right)dy=\left[3y-\frac{y^3}{12}\right]_{-2\sqrt3}^{2\sqrt3}=12\sqrt3-4\sqrt3=8\sqrt3\;.$$

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Darn. 2 minutes too late. :) –  kigen Dec 12 '12 at 5:20
    
@anonymous: Actually, I think that you posted while I was adding the second paragraph, so it’s at worst a tie! –  Brian M. Scott Dec 12 '12 at 5:22
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