Integration?

Question

A curve is defined by parametric equations x=t^2 and y=t^3. A line intersects the curve at t=(-2) and t=3. Find the area enclosed by the line and the curve.

My book says the area is 110. I can get this by integrating the parameters and using t=(-2) and t=3 as the limits. However I could not see why this worked so I found the equation of the line and the curve and then integrated: (the line minus the curve)dy. Using y=(-8) and y=27 as the limits. This time I got 62.5 as the answer.

I can not see why my technique isn't working. It is just a basic integration.

Answer 1

I think you're right, and the book is wrong.

When you find the equation of the curve, you discover that it is the union of y = x^(3/2) and y = -x^(3/2) for x > 0. The line is y = 7x - 36; it intersects y = -x^(3/2) at (4, -8) and y = x^(3/2) at 9, 27). As a result, the enclosed area has y = x^(3/2) as its top boundary on the interval [0, 9], y = -x^(3/2) as its bottom boundary on the interval [0, 4], and y = 7x - 36 as its bottom boundary on the interval [4, 9]. So I broke it up into two integrals like I assume you did, and I found that the area enclosed on [0, 4] was 25.6 and the area enclosed on [4, 9] was 36.9, for a total area of 62.5.