# Thread: Find volume of region, double integrals?

1. ## Find volume of region, double integrals?

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.

2. ## Re: Find volume of region, double integrals?

$y^2 \le x$ implies $-\sqrt{x} \le y \le \sqrt{x}$. Change your bounds of integration.

3. ## Re: Find volume of region, double integrals?

Originally Posted by SlipEternal
$y^2 \le x$ implies $-\sqrt{x} \le y \le \sqrt{x}$. Change your bounds of integration.

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.

4. ## Re: Find volume of region, double integrals?

Thanks, I changed the bounds. And got it correct, but if someone can explain my previous question.

5. ## Re: Find volume of region, double integrals?

The intuition is that in $\displaystyle y^2 \le x$, y can be either positive or negative.

You should probably know from experience that the area $\displaystyle y^2 \le x$ is the whole parabola, while $\displaystyle y \le \sqrt{x}$ is only the upper half. Note that by convention, $\displaystyle \sqrt{x}$ is the positive square root.

Mechanically, when you take the positive square root of both sides of $\displaystyle y^2 \le x$, you get $\displaystyle y \le \sqrt{x}$ if y is positive and $\displaystyle -y \le \sqrt{x}$ if y is negative. The second inequality becomes $\displaystyle y \ge -\sqrt{x}$.

- Hollywood