<p>How do I find the area between two curves in the y-direction?</p>
<p>Between f(y) = y^2 g(y)= y+ 2</p>
<p>when you say in the y-direction do you mean "with respect to y" as in use dy... if so, then the solution is
the integral from -1 to 2 of [(y+2) - (y^2)]dy.</p>
<p>the answer should be: 4.5</p>
<p>Yes thats what I meant. Why is it y+2 y^2 and not the other way around? I cant get it to graph on my calculator in the y-direction. Also, does dy represent the x distance?</p>
<p>I have another question if you dont mind my asking.
When you have to find the area under a curve when the y is less than 0, what do you do?</p>
<p>For example, if you have to find the area between the graph and the curve for f(x)= 2x-x^2 for the interval [0,4], what integral do you write for the part thats under the x-axis (2-4)</p>
<p>it is y+2 first because y+2 is greater than y^2 on that interval. um, to get your calc to graph it, you would have to put everything in terms of x. So since your original equations are x=y+2 and x=y^2, the equations you put into your calculator are y=x-2, y=Sqrt(x) and y=-Sqrt(x). sorry, i will answer the other questions in a minute.</p>
<p>thanks hotpiece.</p>
<p>when the y is less than zero, as in your example, you still take the integral from [0,4]. that will give you the total area, positive and negative combined.</p>
<p>Ok so for the one in my example the area would be the integral from 0 to 2 of (2x-x^2) dx + the integral from 2 to 4 of (2x - x^2) dx?</p>
<p>yes, or you could just do the integral from 0 to 4 of (2x-x^2). you don't have to break them up unless you want to find total area under the curve. in that case, you would take the absolute value of the negative area and add it to the positive area.</p>