Find the area of the region bounded by the graph of the equation 2y^2=x+4 and x=y^2
Did you make a sketch?
It is easy to solve that they intersect at (4,2) and (4,-2)
We also have a nice symmetry so I will take horizontal slices from y=0 to y = 2 , then double the area
Area = 2∫(y^2 - (2y^2 - 4) ) dy from 0 to 2
= 2∫(4 - y^2 ) dy from 0 to 2
= 2[ 4y - (1/3)y^3] | from 0 to 2
= 2[8 - 8/3 - 0 ]
= 2[ 16/3 ]
= 32/3
check my arithmetic
To find the area of the region bounded by the graph of these two equations, we need to follow these steps:
Step 1: Set the two equations equal to each other to find the points of intersection.
2y^2 = x + 4 ......(Equation 1)
x = y^2 ......(Equation 2)
Setting Equation 1 equal to Equation 2:
y^2 = x + 4 ......(Equation 3)
y^2 = y^2 ......(Equation 4)
Since Equation 4 simply states that y^2 is equal to y^2 (which is true for all y), we can conclude that Equation 1 and Equation 2 intersect at all points.
Step 2: Find the points of intersection.
To find the points of intersection, we can set Equation 1 equal to Equation 2 and solve for x and y. Let's set Equation 1 equal to Equation 2 again:
x + 4 = y^2 ......(Equation 5)
Now, rearrange Equation 5 to isolate x:
x = y^2 - 4 ......(Equation 6)
Substitute Equation 6 into Equation 1:
2y^2 = y^2 - 4
Simplifying the equation:
y^2 = -4
Since the square of a real number cannot be negative, there are no real solutions for y. Therefore, there are no points of intersection between the two equations.
Step 3: Analyze the situation.
Since there are no points of intersection, it means that the two equations do not enclose a region. In other words, the region bounded by the graph of the equations does not exist.
Thus, the area of the region bounded by the graph of the equations is zero.