Boundary of the following double integrals (solution check)

Question

I've still been practicing double integrals since my last question and now I have the last 2 left that I am not sure if I did the boundary right and I'd need someone to correct me if I am wrong.

$\left(1\right)$ $$\int \:\int _D\frac{\:x^2}{\sqrt{x^2+y^2}}dxdy\:$$

where$\:D:\:x=0,\:x=y,\:y=1,\:y=\sqrt[3]{2}$

For this one I chose $0\le x\le 1$ so my integral will look like: $$\int _0^1\left(\int _1^{\sqrt[3]{2}}\:\frac{\:x^2}{\sqrt{x^2+y^2}}dy\right)dx$$

$\left(2\right)$ $$\int \:\int _D\:ylnx\:dxdy\:$$ where$\:D:\:xy=1\:,\:y=\sqrt{x},\:x=2$

For this one I really am not sure how to do it honestly, even if I did the graph I don't understand, I assume $0\le x\le 2$ so my integral will be: $$\int _0^2\left(\int _{\frac{1}{x}}^{\sqrt{x}}\:ylnx\:dy\right)dx\:$$

I don't want to evaluate them, just to see if I can do the boundary properly now.

Answer 1

For first one, you will have to split your integral into two if you are integrating over $dy$ first. I would recommend over $dx$ first.

The integral will be,

$\displaystyle \int_1^{2^{1/3}} \int_0^y f(x,y) \ dx \ dy$

For second one, your bounds of $x$ is incorrect. Lower bound is at the intersection of $y = \sqrt x$ and $xy = 1 \implies x = 1$. The upper bound is $x = 2$

So the integral should be,

$ \displaystyle \int_1^2\int _{\frac{1}{x}}^{\sqrt{x}}\: f(x,y) \ dy \ dx\:$

Answer 2

The first integral doesn't seem right, I think you haven't got the correct concept yet. Here is a desmos graph I made:

The line $x=1$, I have brought in just to drill in the point which I am going to explain. Please note that this is not a boundary as given in the question.

The region from the curves in question is the blue area + the green area. For the green area, it just a rectangle so you can do the way you did it. You need to account the blue area as well, which has a bit more trickier integral expression. Here is my desmos graph