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$$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (x^2+y^2+\sin(\pi(x^2+y^2)))\,dy\,dx$$

*sorry if the mathjax is off, I'm new at it.

Anyways, I can use the properties of double integrals to make it

$$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} x^2+y^2 \,dy\,dx + \int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \sin(\pi(x^2+y^2))\,dy\,dx$$

From there I can solve the first double integral expression but I'm not sure on the second double integral expression:

$$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \sin(\pi(x^2+y^2))\,dy\,dx$$

A nudge in the right direction would be appreciated on how to solve this second expression.

Michael Hardy
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user1766888
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2 Answers2

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Convert the integral into polar coordinates. Note that the region you are integrating is the semi-circle of radius lying on the right half plane. Hence, the integral becomes $$\int_{-\pi/2}^{\pi/2} \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr \,d \theta = \pi \cdot \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr$$ All we need to evaluate now is $\displaystyle \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr$. \begin{align} \int_0^3 (r^2 + \sin(\pi r^2)) \,r\, dr & = \int_0^3 r^3 dr + \int_0^3 r \sin( \pi r^2) \,dr = \left. \dfrac{r^4}4 \right \vert_{r=0}^{r=3} + \int_0^3 \dfrac{d( \cos(\pi r^2))}{-2 \pi}\\ & = \dfrac{81}4 - \dfrac{\cos(9 \pi) - \cos(0)}{2 \pi} = \dfrac{81}4 + \dfrac1{\pi} \end{align} Hence, the integral is $$\pi \left(\dfrac{81}4 + \dfrac1{\pi}\right) = 1 + \dfrac{{81}\pi}4$$

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Hint: use polar coordinate i.e. substitute $x=r\cos t$ and $y=r\sin t$

Michael Hardy
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NECing
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