Question

help with multivariable cal!

Answer 1

\[\int\limits_{0}^{1}\int\limits_{0}^{1}\frac{ x }{ 1+xy }dxdy\]
\[\int\limits_{0}^{1}\int\limits_{0}^{1}xy \sqrt{x ^{2}+y^{2}}dydx\]

Answer 2

Mhm I haven't done such stuff yet, but I'd assume you do the inside integral then the outside integral, that's what intuition tells me, maybe we can both learn this :D.

@ganeshie8

Answer 3

Siths! :D

Answer 4

@Astrophysics your intuition is right. We treat the variable we're *not* integrating as a constant. For the first one, we can substitute \(t=1+xy\) so that \(\dfrac{dt}{dx}=y\).
\[\begin{align*}\int\limits_{0}^{1}\int\limits_{0}^{1}\frac{ x }{ 1+xy }~dx~dy&=\int_0^1\int_1^{1+y}\frac{\dfrac{t-1}{y}}{t}\left(\frac{1}{y}~dt\right)~dy\\\\
&=\int_0^1\frac{1}{y^2}\int_1^{1+y}\frac{t-1}{t}~dt~dy
\end{align*}\]
and so on.

Answer 5

For the second one, we would substitute \(p=x^2+y^2\), so that \(\dfrac{dp}{dy}=2y\).
\[\begin{align*}\int\limits_{0}^{1}\int\limits_{0}^{1}xy \sqrt{x ^{2}+y^{2}}~dy~dx&=
\int_0^1\int_{x^2}^{x^2+1}\sqrt p\left(\frac{dp}{2}\right)~dx\\\\
&=\frac{1}{2}\int_0^1\int_{x^2}^{x^2+1}\sqrt p~dp~dx
\end{align*}\]

Answer 6

Generally, we can write
\[\int_a^b\int_c^df(x,y)~dx~dy=\int_a^b\left(\underbrace{\int_c^df(x,y)~dx}_{y\text{ kept constant}}\right)~dy\]

Answer 7

\[\int\limits_{0}^{1}\int\limits_{0}^{1}\frac{ x ^{2} }{ x ^{2} +y ^{2}}dxdy\]

Answer 8

Trig sub: \(x=y\tan z\), so \(dx=y\sec^2z~dz\).
\[\int_{0}^{1}\int_{0}^{\tan^{-1}\frac{1}{y}}\frac{y^2\tan^2z}{y^2\tan^2z+y^2}(y\sec^2z~dz)~dy\\
\int_{0}^{1}y\int_{0}^{\tan^{-1}\frac{1}{y}}\frac{\tan^2z\sec^2z}{\tan^2z+1}~dz~dy\\
\int_{0}^{1}y\int_{0}^{\tan^{-1}\frac{1}{y}}\tan^2z~dz~dy\\
\int_{0}^{1}y\int_{0}^{\tan^{-1}\frac{1}{y}}(\sec^2z-1)~dz~dy
\]

Answer 9

Actually my idea in polar coordinate

Answer 10

Okay, well my suggestion still works, but you're right, polar might make the computation a bit less involved.

First you need to describe the region in polar coordinates: