Video: Finding an Area Enclosed by a Quadratic Equation, the y-Axis, and Two Straight Lines

Find the area enclosed by the graph of π‘₯ = 9 βˆ’ 𝑦², the 𝑦-axis, and the lines 𝑦 = βˆ’3 and 𝑦 = 3.


Video Transcript

Find the area enclosed by the graph of π‘₯ equals nine minus 𝑦 squared, the 𝑦-axis, and the lines 𝑦 equals negative three and 𝑦 equals three.

We notice that, in this example, the area we’ve been asked to find is enclosed by a curve with the equation π‘₯ equals some function of 𝑦. The area is also bounded by the 𝑦-axis rather than the π‘₯-axis and two lines with equations of the form 𝑦 equals some constant, which are horizontal rather than vertical lines.

Now we could repeat the process from first principles to see how we can use integration to find this area, rather than an area bounded by a curve in the form 𝑦 equals 𝑓 of π‘₯ and the π‘₯-axis. But actually, it is as simple as swapping π‘₯ and 𝑦 around. In order to find the area enclosed by a curve in the form π‘₯ equals 𝑔 of 𝑦, the 𝑦-axis, and the two horizontal lines 𝑦 equals 𝑐 and 𝑦 equals 𝑑, we evaluate the definite integral from 𝑐 to 𝑑 of 𝑔 of 𝑦 with respect to 𝑦.

In this case then, we’re evaluating the definite integral from negative three to three of nine minus 𝑦 squared d𝑦. Notice that everything in the integral is in terms of 𝑦, not π‘₯. To integrate powers of 𝑦 not equal to negative one, we increase the power by one and divide by the new power, giving nine 𝑦 minus 𝑦 cubed over three evaluated between negative three and three. We then substitute our limits, giving nine multiplied by three minus three cubed over three minus nine multiplied by negative three minus negative three cubed over three. That’s 27 minus nine minus negative 27 plus nine, which is equal to 36.

So we find that the given area is 36 square units. And in this problem, we’ve seen that, in order to find an area enclosed by the graph of π‘₯ equals some function of 𝑦, the 𝑦-axis, and two horizontal lines, we can just perform a definite integral with everything in terms of 𝑦 rather than everything in terms of π‘₯. Notice also that, in this question, we weren’t concerned that some of the area lay below the π‘₯-axis as this time we were integrating with respect to 𝑦. And all of the area lay on the same side of the 𝑦-axis.

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