0
$\begingroup$

I am trying to calculate the following double integral:

$$\iint_D{dxdy}$$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $\int_{0}^{3} \int_{\frac{4}{3}y}^{\sqrt{25-y^2}} dx dy$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.