IBDP Maths AHL 5.17 Area of the region enclosed by a curve and y axis AA HL Paper 2- Exam Style Questions- New Syllabus

(a)

Since the solid is formed by rotating about the \(y\)-axis, it is convenient to express the curves in terms of \(y\).

From \( y=\sqrt{x^2-16}, \)

we obtain \( x^2=y^2+16. \)

Also, from \( y=2+\frac{x^2}{25}, \)

we obtain \( x^2=25y-50. \)

The required volume is the volume generated by the larger radius minus the volume generated by the smaller radius.

\( V=\pi\int_{0}^{3}(y^2+16)\,dy-\pi\int_{2}^{3}(25y-50)\,dy. \)

Evaluate the first integral:

\( \pi\left[\frac{y^3}{3}+16y\right]_0^3 =\pi(9+48) =57\pi. \)

Evaluate the second integral:

\( \pi\left[\frac{25}{2}y^2-50y\right]_2^3 =\pi\left(\frac{25}{2}\right). \)

Hence,

\( V=57\pi-\frac{25}{2}\pi =\frac{89}{2}\pi. \)

This method uses the washer (disc) technique, where the outer radius is determined by \(f\) and the inner radius by \(g\) over the interval where both curves are present.

✅ Volume \(=\dfrac{89}{2}\pi\) cubic units.