Area of a Region Bounded by Curves
Solved Problems
Example 5.
Find the area of the region enclosed by the curve \(y = \sqrt {x + 1} \) and the line \(y = x + 1.\)
Solution.
It is easy to see that the curve \(y = \sqrt {x + 1} \) and the straight line \(y = x + 1\) intersect at the points \(x = -1,\) \(x = 0\) (Figure \(10\)).
Figure 10. Then
\[A = \int\limits_{ - 1}^0 {\left[ {\sqrt {x + 1} - \left( {x + 1} \right)} \right]dx} = \left. {\frac{{2{{\left( {x + 1} \right)}^{\frac{3}{2}}}}}{3} - \frac{{{x^2}}}{2} - x} \right|_{ - 1}^0 = \left( {\frac{2}{3} - 0 - 0} \right) - \left( {0 - \frac{1}{2} + 1} \right) = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}.\]
Example 6.
Find the area of the region enclosed by the root curve \(y = \sqrt {x} \) and the line \(y = kx,\) where \(k \gt 0.\)
Solution.
First we find the points of intersection of both curves:
\[\sqrt x = kx,\;\; \Rightarrow \sqrt x - kx = 0,\;\; \Rightarrow \sqrt x \left( {1 - k\sqrt x } \right) = 0,\;\; \Rightarrow {x_1} = 0,\;{x_2} = \frac{1}{{{k^2}}}.\]
Figure 11. Now we calculate the area \(A\) using integration:
\[A = \int\limits_0^{\frac{1}{{{k^2}}}} {\left( {\sqrt x - kx} \right)dx} = \left. {\frac{{2{x^{\frac{3}{2}}}}}{3} - \frac{{k{x^2}}}{2}} \right|_0^{\frac{1}{{{k^2}}}} = \frac{2}{3}{\left( {\frac{1}{{{k^2}}}} \right)^{\frac{3}{2}}} - \frac{k}{2}{\left( {\frac{1}{{{k^2}}}} \right)^2} = \frac{2}{{3{k^3}}} - \frac{1}{{2{k^3}}} = \left( {\frac{2}{3} - \frac{1}{2}} \right)\frac{1}{{{k^3}}} = \frac{1}{{6{k^3}}}.\]
Example 7.
Find the area of the region bounded by the curve \(y = {2^x}\) and the lines \(x = 0,\) \(y = 2.\)
Solution.
Figure 12. The upper graph is \(y = 2\) and the lower curve is \(y = {2^x}.\) Hence, the area is
\[A = \int\limits_0^1 {\left( {2 - {2^x}} \right)dx} = \left. {2x - \frac{{{2^x}}}{{\ln 2}}} \right|_0^1 = \left( {2 - \frac{2}{{\ln 2}}} \right) - \left( {0 - \frac{1}{{\ln 2}}} \right) = 2 - \frac{1}{{\ln 2}} \approx 0.56\]
Example 8.
Find the area enclosed by the three petaled rose \(r = \sin 3\theta .\)
Solution.
Figure 13. Since each petal has the same area, we calculate the area of one petal and multiply the result by three. So we have
\[A = \frac{1}{2}\int\limits_0^{\frac{\pi }{3}} {{r^2}\left( \theta \right)d\theta } = \frac{1}{2}\int\limits_0^{\frac{\pi }{3}} {{{\sin }^2}\left( {3\theta } \right)d\theta } = \frac{1}{4}\int\limits_0^{\frac{\pi }{3}} {\left[ {1 - \cos \left( {6\theta } \right)} \right]d\theta } = \frac{1}{4}\left. {\left[ {\theta - \frac{{\sin \left( {6\theta } \right)}}{6}} \right]} \right|_0^{\frac{\pi }{3}} = \frac{1}{4} \cdot \frac{\pi }{3} = \frac{\pi }{{12}}\]
Hence, the area of the all region is \(\frac{\pi }{4}.\)
Example 9.
Find the area enclosed by the cardioid \(r = 1 + \cos \theta .\)
Solution.
Figure 14. We can easily the area of the cardioid by integrating the polar equation in the interval \(\left[ {0,2\pi } \right].\) This yields:
\[A = \frac{1}{2}\int\limits_0^{2\pi } {{r^2}\left( \theta \right)d\theta } = \frac{1}{2}\int\limits_0^{2\pi } {{{\left( {1 + \cos \theta } \right)}^2}d\theta } = \frac{1}{2}\int\limits_0^{2\pi } {\left( {1 + 2\cos \theta + {{\cos }^2}\theta } \right)d\theta } = \frac{1}{2}\int\limits_0^{2\pi } {\left( {1 + 2\cos \theta + \frac{{1 + \cos 2\theta }}{2}} \right)d\theta } = \frac{1}{4}\int\limits_0^{2\pi } {\left( {3 + 4\cos \theta + \cos 2\theta } \right)d\theta } = \frac{1}{4}\left. {\left[ {3\theta + 4\sin \theta + \frac{{\sin 2\theta }}{2}} \right]} \right|_0^{2\pi } = \frac{1}{4} \cdot 6\pi = \frac{{3\pi }}{2}.\]
Example 10.
Find the area of the region bounded by the astroid \({x^{\frac{3}{2}}} + {y^{\frac{3}{2}}} = 1.\)
Solution.
We represent the equation of the astroid in parametric form:
\[x\left( t \right) = {\cos ^3}t,\; y\left( t \right) = {\sin ^3}t.\]
Check by substitution:
\[{x^{\frac{3}{2}}} + {y^{\frac{3}{2}}} = 1,\;\; \Rightarrow {\left( {{{\cos }^3}t} \right)^{\frac{3}{2}}} + {\left( {{{\sin }^3}t} \right)^{\frac{3}{2}}} = 1,\;\; \Rightarrow {\cos ^2}t + {\sin ^2}t = 1,\]
which is true.
Let's now calculate the area of the region enclosed by the parametric curve.
Figure 15. We apply the following integration formula:
\[A = \frac{1}{2}\int\limits_0^T {\left[ {x\left( t \right)y^\prime\left( t \right) - x^\prime\left( t \right)y\left( t \right)} \right]dt} .\]
As
\[x^\prime\left( t \right) = - 3\,{\cos ^2}t\sin t,\;\; y^\prime\left( t \right) = 3\,{\sin ^2}t\cos t,\]
we have
\[A = \frac{1}{2}\int\limits_0^{2\pi } {\left[ {x\left( t \right)y'\left( t \right) - x'\left( t \right)y\left( t \right)} \right]dt} = \frac{1}{2}\int\limits_0^{2\pi } {\left[ {3\,{{\cos }^4}t\,{{\sin }^2}t + 3\,{{\cos }^2}t\,{{\sin }^4}t} \right]dt} = \frac{3}{2}\int\limits_0^{2\pi } {\left[ {{{\cos }^2}t\,{{\sin }^2}t\underbrace {\left( {{{\cos }^2}t + {{\sin }^2}t} \right)}_1} \right]dt} = \frac{3}{8}\int\limits_0^{2\pi } {{{\sin }^2}\left( {2t} \right)dt} = \frac{3}{{16}}\int\limits_0^{2\pi } {\left[ {1 - \cos \left( {4t} \right)} \right]dt} = \frac{3}{{16}}\left. {\left[ {t - \frac{{\sin \left( {4t} \right)}}{4}} \right]} \right|_0^{2\pi } = \frac{3}{{16}} \cdot 2\pi = \frac{{3\pi }}{8}.\]
