definite integralsintegrationmultivariable-calculus
Evaluate the integral $$\iint\limits_D \frac{1}{(x+y+1)^3} \, dA$$ where $D$ is the first quadrant.
In this case, what would the limits of integration be? I'm having trouble moving to polar coordinates. Obviously the unit circle for the first quadrant is the area from $0$ to $\pi/2$, but I'm not sure how to break this up for $x$ and $y$.
You can use polar coordinates, then you have
$$ 9\leq x^2+y^2\leq 25 \implies 3\leq r\leq 5, $$
and
$$ 0\leq \theta \leq \frac{\pi}{4},\quad \rm{since}\quad y=x. $$
$$ \displaystyle\iint\limits_{R}\frac{y^2}{x^2}\ dA = \int_{0}^{\pi/4}\int_{3}^{5}\frac{\sin^{2}(\theta)}{\cos^2(\theta)}rdrd\theta $$
$a)$ One thing you need to know by heart is this transformation: $x^2+y^2=r^2$, so the integrand becomes $\ln (1+r^2)$. We also have $dx\,dy=rdr\,dt$ (I will use $t$ instead of $\theta$ because it is easier for me) To compute the integral in the unit circle, you need to consider not $r=1$ but $r<1$ and $0<t<2\pi$ (they should have taught you this, It would be very difficult for me to explain this here using only text). So our integral is:
$$\int^{2\pi}_0\int^1_0\ln(1+r^2)r\,dr\,dt$$
Substitute $u=r^2$ $\frac{du}{2}=rdr$ and we will get:
$$\int^{2\pi}_0 \frac12 \int^1_0\ln(1+u)\,du\,dt$$
If you integrate by parts, you'll find that the antiderivative of $\ln(u+1)=(u+1) \ln (u+1)-u$, and the whole integral turns out to be:
$$2\pi\ln2-\pi$$
$b)$ I will assume we will find the surface of the Lemniscate. We can divide our function into 4 equal parts, only compute it for the first quadrant, and multiply by 4.
We start by inserting $x=r \cos t \, ,y=r \sin t$ into the equation of the Lemniscate. We will get:
$$r^4=4r^2(\cos ^2t-\sin ^2 t)=4r^2\cos (2t)$$
We want the positive $r$ value so we have $r=2\sqrt{\cos(2t)}$. Hence, $r$ goes from $0$ to $r=2\sqrt{\cos(2t)}$.
We also need the maximum $t$ value. We will need to find the angle of this tangent line:
To find it, we will use implicit derivative of the lemiscate evaluated at $(x,y)=(0,0)$. Which turns out to be $\pm1$. We are concerned with the first quadrant, so we take $+1$. The angle that makes the slope $+1$ is $\arctan(1)=\pi/4$
The surface integral is:
$$4\int^{\pi/4}_0\int^{2\sqrt{\cos(2t)}}_0 r\,dr\,dt$$
I can assume you can take it form here? The answer should come out to be $4$
Best Answer
Hint: Calculate $$\int_{x=0}^\infty\left(\int_{y=0}^\infty \frac{1}{(x+y+1)^3}\,dy\right)\,dx.$$ The inner integral is straightforward, just a power. Same for the second.
Maybe for the first you can let $u=x+y+1$, but it should not be necessary.