© All Rights reserved @ LearnWithDash
Step-by-Step Solution
1) Identify the Curves and Lines Involved
• Parabola: The equation is
2y^2 = 3x , which can be rewritten as
x = \frac{2}{3}y^2 .
This parabola opens to the right and passes through the origin (0,0).
• Line 1: x + y = 3 .
• Line 2: y = 0 (the x-axis).
• Circle: (x-3)^2 + y^2 = 2 , centered at (3,0) with radius \sqrt{2} .
We want the area bounded by these curves but lying outside the given circle.
2) Determine the Region Bounded by Parabola & Lines
The region enclosed before we consider the circle is that bounded by the parabola
x = \frac{2}{3}y^2 , the line x + y = 3 , and the line y = 0 .
Intersection with y=0 :
Substituting y = 0 in 2y^2 = 3x gives x = 0 .
Hence the intersection point is (0,0).
Intersection with x + y = 3 :
Rewrite x + y = 3 as x = 3 - y and substitute into 2y^2 = 3x :
2y^2 = 3(3 - y)
\quad\Longrightarrow\quad
2y^2 + 3y - 9 = 0.
The discriminant is D = 3^2 - 4 \cdot 2 \cdot (-9) = 81, so roots are
y = \frac{-3 \pm 9}{4}.
The positive solution is y = \frac{6}{4} = 1.5, giving x = 1.5.
So they intersect at (1.5, 1.5) .
Another obvious intersection with y=0 and x + y = 3 is (3,0) .
Thus, the region is a curved triangular shape with vertices (0,0), (1.5,1.5), and (3,0).
3) Compute the Area Enclosed by Parabola & Lines (Ignoring the Circle)
Let this region’s area be A' . One way is to integrate w.r.t. y from y=0 to y=1.5 .
For a given y in this range, x goes from
x = \frac{2}{3}y^2
(parabola) to
x = 3 - y
(straight line).
Therefore,
A'
= \int_{y=0}^{1.5}
\Bigl[\bigl(3 - y\bigr) - \tfrac{2}{3}y^2\Bigr]
\, dy.
We split this integral:
\displaystyle \int_{0}^{1.5} (3 - y)\,dy
= \left[3y - \tfrac{y^2}{2}\right]_{0}^{1.5}
= 3\times 1.5 - \frac{(1.5)^2}{2}
= 4.5 - 1.125
= 3.375.
\displaystyle \int_{0}^{1.5} \tfrac{2}{3}y^2\,dy
= \tfrac{2}{3} \int_{0}^{1.5} y^2\,dy
= \tfrac{2}{3} \times \left[\frac{y^3}{3}\right]_{0}^{1.5}
= \tfrac{2}{3} \times \frac{(1.5)^3}{3}
= \tfrac{2}{3} \times \frac{3.375}{3}
= \tfrac{2}{3} \times 1.125
= 0.75.
Hence,
A' = 3.375 \;-\; 0.75
= 2.625
= \frac{21}{8}.
4) Portion of the Region Inside the Circle
We must exclude the part that lies inside the circle
(x-3)^2 + y^2 = 2,
so the desired area is the total A' minus the portion inside the circle.
Intersection with x + y = 3 :
Substitute x = 3 - y into (x-3)^2 + y^2 = 2 :
\bigl((3-y) - 3\bigr)^2 + y^2
= (-y)^2 + y^2 = 2y^2 = 2,
which gives y^2 = 1 (take y = 1 in the first quadrant). Then x=2.
So the circle meets x + y = 3 at (2,1) .
Intersection with y=0 :
(x-3)^2 = 2 \implies x = 3 \pm \sqrt{2}.
In the range x \in [0,3] , we get x = 3 - \sqrt{2} as the intersection on the x-axis.
We see that ≤for 0 ≤ y ≤ 1≥ the circle cuts into the region from
x = 3 - \sqrt{2-y^2}
up to
x = 3 - y.
Hence the area inside the circle (and also inside our original region) is:
A_{\text{inside circle}}
= \int_{0}^{1}
\Bigl[\bigl(3 - y\bigr) - \bigl(3 - \sqrt{2 - y^2}\bigr)\Bigr]
\, dy
= \int_{0}^{1}
\Bigl[\sqrt{2 - y^2} - y\Bigr]
\, dy.
We split it as
\int_{0}^{1}\sqrt{2-y^2}\,dy - \int_{0}^{1}y \,dy.
Integral of \sqrt{2 - y^2} :
Set y = \sqrt{2}\sin\theta , 0 \leq \theta \leq \tfrac{\pi}{4} .
Then \sqrt{2 - y^2} = \sqrt{2}\cos\theta and dy = \sqrt{2}\cos\theta\, d\theta.
Hence
\int_{0}^{1}\sqrt{2 - y^2}\, dy
= \int_{0}^{\pi/4}
\sqrt{2}\cos\theta \times \sqrt{2}\cos\theta\, d\theta
= \int_{0}^{\pi/4} 2\cos^2\theta\, d\theta.
Using 2\cos^2\theta = 1 + \cos(2\theta),
\int_{0}^{\pi/4} 2\cos^2\theta\, d\theta
= \int_{0}^{\pi/4} \bigl(1 + \cos(2\theta)\bigr)\, d\theta
= \left[\theta + \tfrac{1}{2}\sin(2\theta)\right]_{0}^{\pi/4}
= \tfrac{\pi}{4} + \tfrac{1}{2}.
Integral of y from 0 to 1:
\int_{0}^{1} y\, dy
= \left[\frac{y^2}{2}\right]_{0}^{1}
= \frac{1}{2}.
Therefore,
A_{\text{inside circle}}
= \Bigl(\tfrac{\pi}{4} + \tfrac{1}{2}\Bigr) - \frac{1}{2}
= \frac{\pi}{4}.
5) Area Outside the Circle (Required Area A )
Subtract the inside-circle area from A' :
A = A' - A_{\text{inside circle}}
= \frac{21}{8} \;-\; \frac{\pi}{4}
= \frac{21 - 2\pi}{8}.
6) Final Expression: 4\bigl(\pi + 4A\bigr)
First find 4A :
4A
= 4 \times \frac{21 - 2\pi}{8}
= \frac{21 - 2\pi}{2}.
Then add \pi :
\pi + 4A
= \pi + \frac{21 - 2\pi}{2}
= \frac{2\pi + 21 - 2\pi}{2}
= \frac{21}{2}
= 10.5.
Finally multiply by 4:
4\bigl(\pi + 4A\bigr)
= 4 \times 10.5
= 42.
Answer
\displaystyle 42 .
