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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 :

Attachment 30561

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.

Re: Find volume of region, double integrals?

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

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Re: Find volume of region, double integrals?

Quote:

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:

Attachment 30562

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.

Re: Find volume of region, double integrals?

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

Re: Find volume of region, double integrals?

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