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$$\int_{0}^{2\pi} \int_{0}^{1} r^3 sin^2\theta \frac {2}{\sqrt {4-r^2}} dr d\theta$$

Do I need to substitute something?

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You can write it as $$ \left( \int_{0}^{1} \frac{r^{3}}{\sqrt{4 - r^{2}}}dr\right)\left(\int_{0}^{2\pi} \sin^{2}\theta d\theta \right) $$ I believe you can do the second one yourself, and the first one can be done by using the substitution $u = 4-r^{2}$: $$ \int_{0}^{1} \frac{r^{3}}{\sqrt{4-r^{2}}} dr = \int_{4}^{3} \frac{r^{2}}{\sqrt{u}} rdr = \int_{3}^{4} \frac{4-u}{\sqrt{u}}\frac{1}{2} du = \int_{3}^{4} \frac{2}{\sqrt{u}} - \frac{\sqrt{u}}{2}du = \cdots $$

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    $\begingroup$ thanks! i can do next $\endgroup$ – Maggie Jun 9 at 6:45
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Hint:

Try substituting $r = 2 \cos t$ since $(2 \cos t)^2 + (2 \sin t)^2 = 4$.

After this substitution, your integral simplifies to:

$$\int_{0}^{2\pi} \int_{0}^{1} r^3 \sin^2\theta \frac {2}{2 \sin t} (-2 \sin t \ dt) d\theta$$

which cancels, and then rewrite everything in terms of $t$ and $\theta$:

$$-2 \int_{0}^{2\pi} \int_{0}^{1} r^3 \sin^2\theta dt d\theta$$ $$-2 \int_{0}^{2\pi} \sin^2\theta \int_{0}^{1} (2 \cos t)^3 dt d\theta$$ $$-16 \int_{0}^{2\pi} \sin^2\theta \int_{0}^{1} \cos^3 t \ dt d\theta$$

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