I am trying to calculate the following double integral:


from the region:

$$D = \{(x,y) \in R^2 : 0 \le y \le \frac{3}{4}x,\ x^2+y^2\le25 \}$$

So far I have gotten to the point where:

$$\iint_D{dxdy} = \int_0^5\int_{\frac{4}{3}y}^\sqrt{25-y^2}{dxdy}$$

Would that be correct?

  • $\begingroup$ In what point the circle meets the line ? $\endgroup$ – Nosrati Nov 16 '17 at 15:03
  • $\begingroup$ No! It's $(4,3)$, right? $\endgroup$ – Nosrati Nov 16 '17 at 15:15
  • $\begingroup$ $x^2+(\frac34x)^2=25$ $\endgroup$ – Nosrati Nov 16 '17 at 15:16
  • $\begingroup$ Or is it $(4,3)$? $\endgroup$ – Omari Celestine Nov 16 '17 at 15:21
  • $\begingroup$ So your integral is $$\iint_D{dxdy} = \int_0^3\int_{\frac{4}{3}y}^\sqrt{25-y^2}{dxdy}$$ $\endgroup$ – Nosrati Nov 16 '17 at 15:22

Indeed \begin{align} \iint_D{dxdy} &= \int _0^3\int _{\frac{4 y}{3}}^{\sqrt{25-y^2}}1dxdy \\ &= \int _0^4\int _0^{\frac{3 x}{4}}1dydx+\int _4^5\int _0^{\sqrt{25-x^2}}1dydx\\ &= \int _0^{\tan ^{-1}\left(\frac{3}{4}\right)}\int _0^5rdrdt \\ &= \color{blue}{\frac{25}{2} \tan ^{-1}\frac{3}{4}}\\ &\sim 8 \end{align}

  • $\begingroup$ How did you get to the second part with the different intervals? $\endgroup$ – Omari Celestine Nov 16 '17 at 16:00
  • $\begingroup$ Consider region in intervals $[0,4]$ and $[4,5]$. $\endgroup$ – Nosrati Nov 16 '17 at 16:07

Draw the picture. The region is a circular sector, centered at the origin, with radius 5. The sector is in the first quadrant between the $x$-axis and the line $y=\frac{3}{4}x$. Find the intersection of the line and the circle in the first quadrant. It is $(4,3)$.

If you learned double integrals in polar coordinates, then you should get $\int_{0}^{\arctan{3/4}} \int_{0}^{5} r dr d\theta$.

If not polar, then you can get $\int_{0}^{3} \int_{\frac{4}{3}y}^{\sqrt{25-y^2}} dx dy$


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