Question

Use double integration to find the area of the plane region enclosed by the given curves. y² = 324 - x and y² = 324 – 324x.

Answer 1

Firstly we should determine what region do we have.

$y^2=324-x \to y=±\sqrt{324-x}$ . So, graph each of that function is received by tranformation of the graph of the functions $y=±\sqrt{x}$ by inversion relative to y-axis and shifting 324 points to the right

$y^2=324-324x \to y=±\sqrt{324-324x}=±18\sqrt{1-x}$ . So, graph each of that function is received by tranformation of the graph of the functions $y=±\sqrt{x}$ by inversion relative to y-axis, shifting 1 point to the righ and vertical stretch by 18

Picture below represents the given region

The points of interception of the graphs can be found as:

$±18\sqrt{1-x}=±\sqrt{324-x}\to 324-324x= 324-x\to x=0$

$y =±\sqrt{324-0}=±18$

The points is (0,18) and (0,-18)

Since this figure is symmetric, we can find area of the part where y > 0 and multiply it by 2

$\int_{0}^{324}\int_{0}^{\sqrt{324-x}} dydx - \int_{0}^1\int_{0}^{18\sqrt{1-x}} dydx=\int_{0}^{324}\sqrt{324-x} dx-18\int_{0}^{1}\sqrt{1-x} dx=$ 

$=-\int_{0}^{324}\sqrt{324-x} d(324-x)+18\int_{0}^{1}\sqrt{1-x} d(1-x)=-{\frac {2*(324-324)^{1.5}} 3}+{\frac {2*(324-0)^{1.5}} 3}-$

$-12*(1-1)^{1.5}+12*(1-0)^{1.5}=3888-12=3876$

Finally, the area of the region is 2*3876=7752 square units