How can I integrate $\sqrt{1x^2/a^2y^2/b^2} dx dy$ in the ellipse?
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Set $x=ar\cos(t)$ and $br\sin(t)$, where $r \in [0,1]$ and $t \in [0,2\pi)$. We have $$dxdy = \left \vert \begin{bmatrix} a\cos(t) & b\sin(t)\\ ar\sin(t) & br\cos(t) \end{bmatrix}\right \vert dr dt = abrdrdt$$ Hence, the integral \begin{align} I & = {\int\int}_{\text{ellipse}} \sqrt{1\dfrac{x^2}{a^2}\dfrac{y^2}{b^2}}dxdy = \int_0^{2\pi} \int_0^1 \sqrt{1r^2}abrdrdt\\ & = ab \int_0^{2\pi} \int_0^1 r\sqrt{1r^2}drdt = 2\pi ab \int_0^1 r \sqrt{1r^2}dr = \dfrac{2\pi ab}3 \end{align}
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Evzone
Updated on October 05, 2022Comments

Evzone 15 days
I am lookig for $$\int\int _{D} \sqrt{1\frac{x^2}{a^2}\frac{y^2}{b^2}} dxdy $$ where $D$ is defined by $D=\{ (x,y) \in \mathbb{R} \mid \frac{x^2}{a^2} + \frac{y^2}{b^2}\leq 1 \}$.
please help

Evzone over 7 yearsFind the value of double integral sqrt(1x^2/a^2y^2/b^2) in the region bounded by the ellipse.

Hans Lundmark over 7 yearsLet $u=x/a$ and $v=y/b$, then change to polar coordinates.


Evzone over 7 yearsThanks!! you are very good.