Area between two curves

**sj9110** · December 16th 2008, 07:00 PM

I need to find the area between y=sqrt(x+3) and y = (x+3)/2

I integrated in terms of y from 0 to 2 of (y^2 -3) - (2y-3). The answer i got was -4/3, and the actual answer is 4/3. Why is it coming out this way?

Thanks for helping!

**o_O** · December 16th 2008, 07:05 PM

Since we're integrating with respect to y, we have to consider which curve is the 'larger' one horizontally. Sketching both curves yield that $y = \tfrac{1}{2}(x+3)$ is the larger one.

So: $A = \int_0^2 \left[2y - 3 - \left(y^2 - 3\right)\right] dx$

We can see that reversing the order is accounted for by the negative sign.

**sj9110** · December 16th 2008, 07:11 PM

Okay thank you. I understand the concept, and I figured that was the case but I am having a hard time visualizing that 1/2(x+3) is larger. I have the functions graphed out on my paper, when I rotate the graph 90 degree counterclockwise I can see how 1/2(x+3) is larger...is this how to always look at this type of problem when integrating with respect to y?

**o_O** · December 16th 2008, 07:13 PM

Yes. When integrating with respect to x, we find the larger function in the vertical direction (up and down). With respect to y, we look in the horizontal direction (left and right).

**sj9110** · December 16th 2008, 07:19 PM

Okay, thanks again for clarifying.

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