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