Calculate: $\iint_D{dxdy} \text{ from the region: } D = \{(x,y) \in R^2 : 0 \le y \le \frac{3}{4}x,\ x^2+y^2\le25 \}$

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?

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

• Consider region in intervals $[0,4]$ and $[4,5]$. – 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 $\int_{0}^{3} \int_{\frac{4}{3}y}^{\sqrt{25-y^2}} dx dy$