1. Limits of Integration

Can I get some help, im trying to view the image of this in space

r is the region in the first quadrant bounded by the line y=x and
the curve x^4 +x^2y^2=y^2..
any idea on the limits for my integral, i think it goes from 0-pi/2 and something
else

2. I don't think changing to polar coordinates helps. $\displaystyle x^2y^2- y^2= y^2(x^- 1)= -x^4$ so $\displaystyle y^2= \frac{x^4}{1- x^2}$ and, in the first quadrant, $\displaystyle y= \frac{x^2}{\sqrt{1- x^2}}$. That intersects y= x when x= 0 and when $\displaystyle \frac{x^4}{1- x^2}= x^2$ which is equivalent to $\displaystyle x^2= 1- x^2$.

3. (double integral)

well my integral is
(1+x^2 +y^2)^-2
i think polar coordinates is appropriate.
but i think 0-pie/2 for theta but what is the limits for r dr?
clearly you have solved for y, but im not visualizing the line y=x and the curve stated above

4. Hello, begin!

$\displaystyle R\text{ is the region in the first quadrant bounded by the line }y=x$$\displaystyle \text{and the curve } x^4 +x^2y^2\:=\:y^2$

I would convert to polar coordinates . . .

We have: .$\displaystyle x^2(x^2+y^2) \:=\:y^2$

Then: .$\displaystyle r^2\cos^2\!\theta \cdot r^2 \:=\:r^2\sin^2\!\theta \quad\Rightarrow\quad r^2 \:=\:\dfrac{\sin^2\!\theta}{\cos^2\!\theta} \quad\Rightarrow\quad r^2 \:=\:\tan^2\theta$

Area Formula: .$\displaystyle \displaystyle A \;=\;\tfrac{1}{2}\int^{\beta}_{\alpha} \!r^2\,d\theta$

We have: .$\displaystyle r^2 \,=\,\tan^2\!\theta,\;\alpha = 0,\;\beta = \frac{\pi}{4}$

Go for it!

5. i have theta goes from 0-pie/2 but what abour r dr