$y^2 \le x$ implies $-\sqrt{x} \le y \le \sqrt{x}$. Change your bounds of integration.
Third semester calculus; What I'm I doing wrong? I'm being mark wrong for the answer I got.
My problem:
Find the volume of the region under the graph of f(x,y)= 5x+y+1 and above the region y^2<=x, 0<=x<=16.
My attempt:
First I drew the diagram for my bounds :
From here I setup the iterated integral (double integral) with the bounds I found ( wolframalpha link ):
int (5x+y+1)dydx, x=0 to 16, y=0 to sqrt(x) - Wolfram|Alpha
First I integrate with respect to y by holding x constant, then I integrate with respect to x and get the answer of 6464/3 cubic units.
So you're saying that the drawing should be like so:
I don't understand the intuition behind that? So when we take the square root of y^2 <= x we obtain y<= sqrt(x) and y => -sqrt(x) ? For the latter equation the inequality sign flipped because we introduced a negative sign due to the square root being positive and negative? I think my mistake was thinking that the x-axis was bounded with y=0.
The intuition is that in , y can be either positive or negative.
You should probably know from experience that the area is the whole parabola, while is only the upper half. Note that by convention, is the positive square root.
Mechanically, when you take the positive square root of both sides of , you get if y is positive and if y is negative. The second inequality becomes .
- Hollywood