This note shows how to find areas between a curve and the y-axis, and how to find the volume of a solid made by spinning a region about the x-axis or the y-axis. Both ideas turn a picture into a definite integral, so they are a regular source of longer calculus questions.
Key ideas
Normally we add up thin vertical strips, with width \(dx\). For an area next to the y-axis it is easier to add up thin horizontal strips, with height \(dy\). Each strip has length \(x\) (the distance from the y-axis to the curve) and thickness \(dy\), so the area is found by integrating \(x\) with respect to \(y\).
- Rewrite the curve as \(x = g(y)\) if it is given as \(y = f(x)\). This may need an inverse function, for example \(y=\ln x\) becomes \(x=e^{y}\).
- The limits are values of \(y\), not values of \(x\). Read them from the question or find them by substituting the \(x\)-boundaries into the curve.
- If the curve lies to the left of the y-axis, \(x\) is negative and the integral gives a negative value. The area is the absolute value.
Volumes of revolution. When a region is rotated through \(360^\circ\) (one full turn) about an axis, every thin strip sweeps out a thin disc.
- About the x-axis: each disc has radius \(y\) and thickness \(dx\). Its volume is \(\pi y^{2}\,dx\).
- About the y-axis: each disc has radius \(x\) and thickness \(dy\). Its volume is \(\pi x^{2}\,dy\).
Add the discs with an integral. The variable inside the integral must match the \(dx\) or \(dy\). For a rotation about the y-axis, write \(x^{2}\) in terms of \(y\) before integrating.
Always square the whole expression for the radius first, then integrate. For example, if \(y = 2+x\), the integrand is \((2+x)^{2}\), not \(4+x^{2}\).
Why the formulas work. A disc of radius \(r\) and thickness \(t\) has volume \(\pi r^{2}t\). As the thickness shrinks to \(dx\) or \(dy\), adding the discs becomes an integral. The same idea applies to area: a strip of length \(x\) and thickness \(dy\) has area \(x\,dy\).
Choosing the method. If the curve is easy to rearrange as \(x=g(y)\), integrate with \(dy\). If it is not, you can sometimes find the same area as a rectangle minus an area under the curve, using \(\int y\,dx\). Both give the same value. Check which route has the simpler algebra before you start.
Units. Areas are in square units and volumes in cubic units. If lengths are in centimetres, the volume is in cm\(^{3}\). A calculator can check a definite integral, using its numerical integration feature, but show the working in full unless told otherwise.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(A=\int_{c}^{d} x\,dy\) | Area between the curve \(x=g(y)\), the y-axis and the lines \(y=c\) and \(y=d\), when \(g(y)\ge 0\). |
| \(V=\pi\int_{a}^{b} y^{2}\,dx\) | Volume when the region under \(y=f(x)\) between \(x=a\) and \(x=b\) is rotated \(360^\circ\) about the x-axis. |
| \(V=\pi\int_{c}^{d} x^{2}\,dy\) | Volume when the region between \(x=g(y)\) and the y-axis, from \(y=c\) to \(y=d\), is rotated \(360^\circ\) about the y-axis. |
| Solid of revolution | A 3D shape formed by rotating a region about a straight line. |
| Inverse of \(y=f(x)\) | Needed to write \(x\) in terms of \(y\). Swap the roles of \(x\) and \(y\) and solve. |
Worked examples
Example 1 The curve \(y=\ln x\) is drawn for \(x\gt 0\). Find the exact area of the region bounded by the curve, the y-axis and the lines \(y=1\) and \(y=3\).
Make \(x\) the subject: \(y=\ln x\) gives \(x=e^{y}\).
The limits are already in terms of \(y\): from \(y=1\) to \(y=3\).
\[A=\int_{1}^{3} e^{y}\,dy=\left[e^{y}\right]_{1}^{3}=e^{3}-e\]
Check with a calculator: \(e^{3}-e\approx 20.0855-2.7183=17.4\).
Answer: \(e^{3}-e\) square units (about 17.4).
Example 2 A designer models the profile of a vase by the curve \(y=\sqrt{4x+1}\) (lengths in cm) for \(0\le x\le 2\). The region between the curve, the x-axis, the y-axis and the line \(x=2\) is rotated \(360^\circ\) about the x-axis. Find the volume of the vase.
The radius of each disc is \(y\), so \(y^{2}=4x+1\).
\[V=\pi\int_{0}^{2}(4x+1)\,dx=\pi\left[2x^{2}+x\right]_{0}^{2}\]
At \(x=2\): \(2(4)+2=10\). At \(x=0\): \(0\).
\[V=10\pi\approx 31.4\]
Answer: \(10\pi\approx 31.4\ \text{cm}^{3}\).
Example 3 A bowl is made by rotating the curve \(y=\tfrac12 x^{2}\), for \(0\le y\le 8\), through \(360^\circ\) about the y-axis (lengths in cm). Find the volume of the bowl.
The rotation is about the y-axis, so we need \(x^{2}\) in terms of \(y\).
From \(y=\tfrac12 x^{2}\) we get \(x^{2}=2y\).
\[V=\pi\int_{0}^{8} 2y\,dy=\pi\left[y^{2}\right]_{0}^{8}=\pi(64-0)\]
\[V=64\pi\approx 201\]
Answer: \(64\pi\approx 201\ \text{cm}^{3}\).
Common errors
- Common error: Using \(x\)-limits in an integral with \(dy\). The limits must be \(y\)-values. If the question gives \(x\)-values, substitute them into the curve to get the matching \(y\)-values.
- Common error: Forgetting to square. The volume integrand is \(y^{2}\) (or \(x^{2}\)), not \(y\).
- Common error: Leaving out \(\pi\), or putting \(\pi\) inside the brackets and then squaring it.
- Common error: Integrating \(y^{2}\) with respect to \(x\) without first substituting \(y=f(x)\) (or the other way round for the y-axis). Everything inside the integral must be in terms of one variable.
- Common error: Expanding \((f(x))^{2}\) wrongly, for example squaring each term separately.
Exam tip: Sketch the region and mark the axis of rotation before you write any integral. Then choose \(dx\) for the x-axis and \(dy\) for the y-axis, and give an exact answer in terms of \(\pi\) unless the question asks for a decimal.
Check yourself
- Find the area bounded by the curve \(x=y^{2}+1\), the y-axis and the lines \(y=0\) and \(y=3\).
Answer
\(A=\int_{0}^{3}(y^{2}+1)\,dy=\left[\tfrac{y^{3}}{3}+y\right]_{0}^{3}=9+3=12\) square units. - The region under \(y=e^{x}\) from \(x=0\) to \(x=1\) is rotated \(360^\circ\) about the x-axis. Find the volume to 3 s.f.
Answer
\(V=\pi\int_{0}^{1}e^{2x}\,dx=\pi\left[\tfrac12 e^{2x}\right]_{0}^{1}=\tfrac{\pi}{2}(e^{2}-1)\approx 10.0\) cubic units. - The region between the curve \(y=x^{2}\) and the y-axis, for \(0\le y\le 9\), is rotated \(360^\circ\) about the y-axis. Find the exact volume.
Answer
\(x^{2}=y\), so \(V=\pi\int_{0}^{9}y\,dy=\pi\left[\tfrac{y^{2}}{2}\right]_{0}^{9}=\tfrac{81\pi}{2}\) cubic units. - Find the area between the y-axis and the curve \(x=4-y^{2}\) for \(0\le y\le 2\).
Answer
\(A=\int_{0}^{2}(4-y^{2})\,dy=\left[4y-\tfrac{y^{3}}{3}\right]_{0}^{2}=8-\tfrac{8}{3}=\tfrac{16}{3}\approx 5.33\) square units.