Question Video: Finding the Area of a Region Bounded by Two Quadratic Functions Defined with Respect to 𝑦 Mathematics

Video Transcript

Find the area of the region bounded by 𝑥 is equal to negative five 𝑦 squared plus one and 𝑥 is equal to two 𝑦 squared minus five.

In this question, we’re asked to find the area of a region bounded by two curves: 𝑥 is equal to negative five 𝑦 squared plus one and 𝑥 is equal to two 𝑦 squared minus five. And there’s something interesting we can note about the two curves we’re given. Instead of being given 𝑦 as a function in 𝑥, we’re given 𝑥 as a function in 𝑦. In particular, we’re given quadratics; 𝑥 is a quadratic in 𝑦. And whenever we’re asked to calculate the area of a region bounded by two curves, it’s always a good idea to sketch the two curves we’re given.

And to sketch these two curves, since they’re quadratics, we can use any of our rules or properties of quadratics to help us sketch the curves. However, we are going to need to change these results slightly, since 𝑥 is now a function in 𝑦. Let’s start by sketching our first curve. There’s a few different ways we could do this. One way is to note that we’re given the vertex form of this curve. We see if we substitute 𝑦 is equal to zero into our function, we get one. So the vertex of this curve is one, zero. This is the opposite way round it would be if we were given 𝑦 as a function of 𝑥. So we do need to be careful that we get the coordinates the correct way round.

So we can add this onto our diagram where we won’t include the coordinates, since it’s not necessary. We know that the shape of this curve will be parabolic, since it’s a quadratic curve. However, to help us sketch this, let’s find the 𝑦-intercepts of this curve. We substitute 𝑥 is equal to zero into the curve to get zero is equal to negative five 𝑦 squared plus one.

If we solve this equation for 𝑦, we’ll find the 𝑦-intercepts of the curve. We can solve this equation for 𝑦 by subtracting one from both sides of the equation to get negative one is equal to negative five 𝑦 squared. We divide through by negative five to get one-fifth is equal to 𝑦 squared. And then we take the square root of both sides of the equation, where we know we’ll get a positive and negative root. 𝑦 is equal to positive or negative one over root five. We can add both of these 𝑦-intercepts into our diagram.

And now we can just connect the three points of a parabolic curve to sketch 𝑥 is equal to negative five 𝑦 squared plus one, where since the 𝑥-values of this curve are the outputs of our function, we can see the coordinates of our vertex is a maximum.

Let’s now apply exactly the same process to our second curve. Once again, our curve is given in vertex form. If we substitute 𝑦 is equal to zero into the equation of this curve, we get 𝑥 is equal to negative five. So the vertex of this curve is the point negative five, zero. We can add this point onto our diagram. Now to help us sketch this curve, let’s find the 𝑦-intercepts of the curve.

Once again, we can find the 𝑦-intercepts of the curve by setting 𝑥 is equal to zero and solving the resulting equation, zero is equal to two 𝑦 squared minus five. We add five to both sides of the equation to get five is equal to two 𝑦 squared, divide the equation through by two to get five over two is equal to 𝑦 squared, and then we take the square root of both sides of the equation, where we know we get a positive and negative root. 𝑦 is equal to positive or negative root five over two. These are the 𝑦-intercepts of this curve.

And it’s worth pointing out root five over two is bigger than one over root five. So the 𝑦-intercepts of this curve will be further from the origin than our previous curve. Now all we need to do is connect these three points with a parabolic shape to sketch 𝑥 is equal to two 𝑦 squared minus five. And remember, since the 𝑥-coordinates are the outputs of this function, we can see our vertex is now a minimum value. It’s the furthest point to the left of this curve. And we can now shade in the region bounded by the two curves.

To find this area, we need to notice something interesting about the shaded region. Since the 𝑥-coordinates are the outputs of this function, the further to the right, the higher the output. So we can see this region is bounded above by the curve in pink and bounded below by the curve in orange, since the curve in pink is to the right of the curve in orange over this region. This means we can find this area by using one of our integral results, where we’ll need to rewrite this to be in terms of 𝑦 instead of 𝑥.

If we have two integrable functions 𝑓 of 𝑦 and 𝑔 of 𝑦, where 𝑓 of 𝑦 is greater than or equal to 𝑔 of 𝑦 for 𝑦 over some closed interval from 𝑎 to 𝑏, then the area of the region bounded by the curves 𝑥 is equal to 𝑓 of 𝑦 and 𝑥 is equal to 𝑔 of 𝑦 and the horizontal lines 𝑦 is equal to 𝑎 and 𝑦 is equal to 𝑏 is given by the definite integral from 𝑎 to 𝑏 of 𝑓 of 𝑦 minus 𝑔 of 𝑦 with respect to 𝑦. And we can apply this to find the area of the region in our diagram.

We’ve already shown that the pink curve lies above the orange curve over this region, so 𝑓 of 𝑦 will be negative five 𝑦 squared plus one, and 𝑔 of 𝑦 will be two 𝑦 squared minus five. Next, we know both of these functions are integrable, because they’re polynomials. So they’re integrable over the entire set of real numbers. Finally, we can set our values of 𝑏 and 𝑎 to be the 𝑦-coordinates of the points of intersection between the two curves. Then the region is bounded by these two curves and these two horizontal lines, and so we can use the definite integral to determine its area.

This means we now need to find the values of the 𝑦-coordinates of the points of intersection between the two curves, the values of 𝑎 and 𝑏. To find the 𝑦-coordinates of the points of intersection, we need to set the two functions equal to each other. We need to solve the equation negative five 𝑦 squared plus one is equal to two 𝑦 squared minus five. We can solve this equation by adding five to both sides of the equation. And adding five 𝑦 squared to both sides of the equation, we get six is equal to seven 𝑦 squared. We then divide through by seven to get six over seven is equal to 𝑦 squared. Then we take the square root of both sides of the equation. We get 𝑦 is equal to positive or negative the square root of six over seven. And since 𝑏 is bigger than 𝑎, 𝑏 is positive root six over seven and 𝑎 is negative root six over seven.

We can now substitute all of this information into our integral result to determine the area of the shaded region. It’s equal to the definite integral from negative root six over seven to root six over seven of negative five 𝑦 squared plus one minus two 𝑦 squared minus five with respect to 𝑦.

All that’s left to do now is evaluate this integral to determine the area. First, let’s simplify our integrand. We can distribute the negative over the parentheses and simplify. We get negative seven 𝑦 squared plus six. This is any integral of a polynomial. And we can do this by using the power rule for integration. We need to add one onto the exponents of 𝑦 and then divide by these new exponents. We get negative seven 𝑦 cubed over three plus six 𝑦 evaluated at limits of integration, 𝑦 is equal to negative root six over seven and 𝑦 is equal to root six over seven.

We could now find the difference in the antiderivative evaluated at the upper and lower limits of integration. This would work; however, it would give us a very complicated expression. Instead, let’s simplify this process slightly by noticing something interesting about our antiderivative. We know that 𝑦 cubed is an odd function, so negative seven 𝑦 cubed over three is an odd function, and six 𝑦 is also an odd function. Therefore, the sum of these two values is an odd function. The antiderivative is an odd function. We can use this to simplify the evaluation of our antiderivative.

If ℎ is any odd function, then ℎ evaluated at 𝑐 minus ℎ evaluated at negative 𝑐 is equal to ℎ evaluated at 𝑐 plus ℎ evaluated at 𝑐. It’s two ℎ of 𝑐. And this is exactly what we would be calculating if we found the difference in this antiderivative evaluated at the upper and lower limits of integration, since these are the same value with opposite signs. Therefore, by using this property, we can just double the antiderivative evaluated at the upper limit of integration. We get two multiplied by negative seven times root six over seven cubed over three plus six times root six over seven.

If we then evaluate this expression, we get eight root 42 over seven, which is our final answer. And it’s also worth noting since this represents an area, we could say that this is eight root 42 over seven square units. Therefore, we were able to show the area of the region bounded by the curves 𝑥 is equal to negative five 𝑦 squared plus one and 𝑥 is equal to two 𝑦 squared minus five is eight root 42 over seven.