86981
The area is square units bounded by the normal at \((1,2)\) to parabola \(y^{2}=4 x, x\)-axis and the curve is given by
1 \(\frac{10}{3}\)
2 \(\frac{7}{3}\)
3 \(\frac{4}{3}\)
4 None of these
Explanation:
(A) : Given, \(\mathrm{y}^{2}=4 \mathrm{x}\) Differentiating with respect to \(\mathrm{x}\), \(2 y \frac{d y}{d x}=4\) \(\frac{d y}{d x}=\frac{2}{y}\) \(\left(\frac{d y}{d x}\right)_{(1,2)}=1\) Equation to normal to the curve at \((1,2)\), \(y-y_{1}=\frac{1}{\left(-\frac{d y}{d x}\right)}\left(x-x_{1}\right)\) \(\Rightarrow(\mathrm{y}-2)=-\frac{1}{1}(\mathrm{x}-1) \Rightarrow \mathrm{y}-2=-\mathrm{x}+1 \Rightarrow \mathrm{x}+\mathrm{y}=3\) Line \(\mathrm{x}+\mathrm{y}=3\) meet the \(\mathrm{x}-\) axis at \(\mathrm{x}=3\) Now required area, \(A=\int_{0}^{1} \sqrt{4 x} \cdot d x+\int_{1}^{3}(3-x) \cdot d x\) \(A=2\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1}+\left[3 x-\frac{x^{2}}{2}\right]_{1}^{3}\) \(\mathrm{A}=\frac{4}{3}+\left[9-\frac{9}{2}-3+\frac{1}{2}\right]\) \(\mathrm{A}=\frac{4}{3}+\left(\frac{9}{2}-\frac{5}{2}\right)=\frac{4}{3}+\frac{4}{2}=\frac{20}{6} \Rightarrow \mathrm{A}=\frac{10}{3}\) square units.
COMEDK-2015
Application of the Integrals
86982
Area of the triangle formed by the line \(x+y=3\) and angle bisectors of the pair of straight lines \(x^{2}-y^{2}+2 y-1\) is
1 2 sq. units
2 4 sq. units
3 6 sq. units
4 8 sq. units
Explanation:
(A) : \(\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{y}=1 \Rightarrow \mathrm{x}= \pm(\mathrm{y}-1)\) Bisectors of above line are \(x=0 \& y=1\) So area between \(\mathrm{x}=0, \mathrm{y}=1 \& \mathrm{x}+\mathrm{y}=3\) is shaded Region shown in figure. Area \(=\frac{1}{2} \times 2 \times 2=2\) sq. units
BITSAT-2009
Application of the Integrals
86991
Area bounded by parabola \(y^{2}=x\) and straight line \(2 y=x\) is
1 \(4 / 3\)
2 1
3 \(2 / 3\)
4 \(1 / 3\)
Explanation:
(A) : Parabola \(\mathrm{y}^{2}=\mathrm{x}\) and straight line \(2 \mathrm{y}=\mathrm{x}\) Point of intersection \(\left(\frac{x}{2}\right)^{2}=x\) \(x=0,4\) \(y=0,2\) Area \(\int_{0}^{4} \sqrt{x d x}-\int_{0}^{4}\left(x^{1 / 2}\right) d x\) \(=\frac{2}{3}\left[x^{2 / 3}\right]_{0}^{4}-\left[\frac{x^{2}}{4}\right]_{0}^{4}=\frac{2 \times 8}{3}-4=\frac{4}{3}\)
AMU-2002
Application of the Integrals
86985
What is the area of a loop of the curve \(r=\) \(\operatorname{asin} 3 \theta\) ?
1 \(\frac{\pi \mathrm{a}^{2}}{6}\)
2 \(\frac{\pi \mathrm{a}^{2}}{8}\)
3 \(\frac{\pi \mathrm{a}^{2}}{12}\)
4 \(\frac{\pi \mathrm{a}^{2}}{24}\)
Explanation:
(D) : If curve \(\mathrm{r}=\mathrm{a} \sin 3 \theta\) To trace the curve, we consider the following Table : | $3 \theta=$ | 0 | $\frac{\pi}{2}$ | $\pi$ | $\frac{3 \pi}{2}$ | $2 \pi$ | $\frac{5 \pi}{2}$ | $3 \pi$ | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $\theta=$ | 0 | $\frac{\pi}{6}$ | $\frac{\pi}{3}$ | $\frac{\pi}{2}$ | $\frac{2 \pi}{3}$ | $\frac{5 \pi}{6}$ | $\pi$ | | $\mathrm{r}=$ | 0 | $\mathrm{a}$ | 0 | $-\mathrm{a}$ | 0 | $\mathrm{a}$ | 0 | Thus there is a looping between \(\theta=0 \& \theta=\frac{\pi}{3}\) as \(\mathrm{r}\) varies from \(r=0\) to \(r=0\). Hence, the area of the loop lying in the positive quadrant \(=\frac{1}{2} \int_{0}^{\frac{\pi}{3}} \mathrm{r}^{2} \mathrm{~d} \theta=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin ^{2} \phi \cdot \frac{1}{3} \mathrm{~d} \phi\) [On putting, \(3 \theta=\phi \Rightarrow \mathrm{d} \theta=\frac{1}{3} \mathrm{~d} \phi\) ] \(=\frac{\mathrm{a}^{2}}{6} \int_{0}^{\frac{\pi}{2}} \sin ^{2} \phi \mathrm{d} \phi\) \(=\frac{\mathrm{a}^{2}}{6} \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 \phi}{2} \mathrm{~d} \theta\left[\because \cos 2 \theta=1-2 \sin ^{2} \theta\right]\) \(=\frac{\mathrm{a}^{2}}{12} \cdot\left[\phi+\frac{\sin 2 \phi}{2}\right]_{0}^{\frac{\pi}{2}}=\frac{\mathrm{a}^{2}}{12}\left[\frac{\pi}{2}+\sin \pi\right]=\frac{\pi \mathrm{a}^{2}}{24}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
86981
The area is square units bounded by the normal at \((1,2)\) to parabola \(y^{2}=4 x, x\)-axis and the curve is given by
1 \(\frac{10}{3}\)
2 \(\frac{7}{3}\)
3 \(\frac{4}{3}\)
4 None of these
Explanation:
(A) : Given, \(\mathrm{y}^{2}=4 \mathrm{x}\) Differentiating with respect to \(\mathrm{x}\), \(2 y \frac{d y}{d x}=4\) \(\frac{d y}{d x}=\frac{2}{y}\) \(\left(\frac{d y}{d x}\right)_{(1,2)}=1\) Equation to normal to the curve at \((1,2)\), \(y-y_{1}=\frac{1}{\left(-\frac{d y}{d x}\right)}\left(x-x_{1}\right)\) \(\Rightarrow(\mathrm{y}-2)=-\frac{1}{1}(\mathrm{x}-1) \Rightarrow \mathrm{y}-2=-\mathrm{x}+1 \Rightarrow \mathrm{x}+\mathrm{y}=3\) Line \(\mathrm{x}+\mathrm{y}=3\) meet the \(\mathrm{x}-\) axis at \(\mathrm{x}=3\) Now required area, \(A=\int_{0}^{1} \sqrt{4 x} \cdot d x+\int_{1}^{3}(3-x) \cdot d x\) \(A=2\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1}+\left[3 x-\frac{x^{2}}{2}\right]_{1}^{3}\) \(\mathrm{A}=\frac{4}{3}+\left[9-\frac{9}{2}-3+\frac{1}{2}\right]\) \(\mathrm{A}=\frac{4}{3}+\left(\frac{9}{2}-\frac{5}{2}\right)=\frac{4}{3}+\frac{4}{2}=\frac{20}{6} \Rightarrow \mathrm{A}=\frac{10}{3}\) square units.
COMEDK-2015
Application of the Integrals
86982
Area of the triangle formed by the line \(x+y=3\) and angle bisectors of the pair of straight lines \(x^{2}-y^{2}+2 y-1\) is
1 2 sq. units
2 4 sq. units
3 6 sq. units
4 8 sq. units
Explanation:
(A) : \(\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{y}=1 \Rightarrow \mathrm{x}= \pm(\mathrm{y}-1)\) Bisectors of above line are \(x=0 \& y=1\) So area between \(\mathrm{x}=0, \mathrm{y}=1 \& \mathrm{x}+\mathrm{y}=3\) is shaded Region shown in figure. Area \(=\frac{1}{2} \times 2 \times 2=2\) sq. units
BITSAT-2009
Application of the Integrals
86991
Area bounded by parabola \(y^{2}=x\) and straight line \(2 y=x\) is
1 \(4 / 3\)
2 1
3 \(2 / 3\)
4 \(1 / 3\)
Explanation:
(A) : Parabola \(\mathrm{y}^{2}=\mathrm{x}\) and straight line \(2 \mathrm{y}=\mathrm{x}\) Point of intersection \(\left(\frac{x}{2}\right)^{2}=x\) \(x=0,4\) \(y=0,2\) Area \(\int_{0}^{4} \sqrt{x d x}-\int_{0}^{4}\left(x^{1 / 2}\right) d x\) \(=\frac{2}{3}\left[x^{2 / 3}\right]_{0}^{4}-\left[\frac{x^{2}}{4}\right]_{0}^{4}=\frac{2 \times 8}{3}-4=\frac{4}{3}\)
AMU-2002
Application of the Integrals
86985
What is the area of a loop of the curve \(r=\) \(\operatorname{asin} 3 \theta\) ?
1 \(\frac{\pi \mathrm{a}^{2}}{6}\)
2 \(\frac{\pi \mathrm{a}^{2}}{8}\)
3 \(\frac{\pi \mathrm{a}^{2}}{12}\)
4 \(\frac{\pi \mathrm{a}^{2}}{24}\)
Explanation:
(D) : If curve \(\mathrm{r}=\mathrm{a} \sin 3 \theta\) To trace the curve, we consider the following Table : | $3 \theta=$ | 0 | $\frac{\pi}{2}$ | $\pi$ | $\frac{3 \pi}{2}$ | $2 \pi$ | $\frac{5 \pi}{2}$ | $3 \pi$ | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $\theta=$ | 0 | $\frac{\pi}{6}$ | $\frac{\pi}{3}$ | $\frac{\pi}{2}$ | $\frac{2 \pi}{3}$ | $\frac{5 \pi}{6}$ | $\pi$ | | $\mathrm{r}=$ | 0 | $\mathrm{a}$ | 0 | $-\mathrm{a}$ | 0 | $\mathrm{a}$ | 0 | Thus there is a looping between \(\theta=0 \& \theta=\frac{\pi}{3}\) as \(\mathrm{r}\) varies from \(r=0\) to \(r=0\). Hence, the area of the loop lying in the positive quadrant \(=\frac{1}{2} \int_{0}^{\frac{\pi}{3}} \mathrm{r}^{2} \mathrm{~d} \theta=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin ^{2} \phi \cdot \frac{1}{3} \mathrm{~d} \phi\) [On putting, \(3 \theta=\phi \Rightarrow \mathrm{d} \theta=\frac{1}{3} \mathrm{~d} \phi\) ] \(=\frac{\mathrm{a}^{2}}{6} \int_{0}^{\frac{\pi}{2}} \sin ^{2} \phi \mathrm{d} \phi\) \(=\frac{\mathrm{a}^{2}}{6} \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 \phi}{2} \mathrm{~d} \theta\left[\because \cos 2 \theta=1-2 \sin ^{2} \theta\right]\) \(=\frac{\mathrm{a}^{2}}{12} \cdot\left[\phi+\frac{\sin 2 \phi}{2}\right]_{0}^{\frac{\pi}{2}}=\frac{\mathrm{a}^{2}}{12}\left[\frac{\pi}{2}+\sin \pi\right]=\frac{\pi \mathrm{a}^{2}}{24}\)
86981
The area is square units bounded by the normal at \((1,2)\) to parabola \(y^{2}=4 x, x\)-axis and the curve is given by
1 \(\frac{10}{3}\)
2 \(\frac{7}{3}\)
3 \(\frac{4}{3}\)
4 None of these
Explanation:
(A) : Given, \(\mathrm{y}^{2}=4 \mathrm{x}\) Differentiating with respect to \(\mathrm{x}\), \(2 y \frac{d y}{d x}=4\) \(\frac{d y}{d x}=\frac{2}{y}\) \(\left(\frac{d y}{d x}\right)_{(1,2)}=1\) Equation to normal to the curve at \((1,2)\), \(y-y_{1}=\frac{1}{\left(-\frac{d y}{d x}\right)}\left(x-x_{1}\right)\) \(\Rightarrow(\mathrm{y}-2)=-\frac{1}{1}(\mathrm{x}-1) \Rightarrow \mathrm{y}-2=-\mathrm{x}+1 \Rightarrow \mathrm{x}+\mathrm{y}=3\) Line \(\mathrm{x}+\mathrm{y}=3\) meet the \(\mathrm{x}-\) axis at \(\mathrm{x}=3\) Now required area, \(A=\int_{0}^{1} \sqrt{4 x} \cdot d x+\int_{1}^{3}(3-x) \cdot d x\) \(A=2\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1}+\left[3 x-\frac{x^{2}}{2}\right]_{1}^{3}\) \(\mathrm{A}=\frac{4}{3}+\left[9-\frac{9}{2}-3+\frac{1}{2}\right]\) \(\mathrm{A}=\frac{4}{3}+\left(\frac{9}{2}-\frac{5}{2}\right)=\frac{4}{3}+\frac{4}{2}=\frac{20}{6} \Rightarrow \mathrm{A}=\frac{10}{3}\) square units.
COMEDK-2015
Application of the Integrals
86982
Area of the triangle formed by the line \(x+y=3\) and angle bisectors of the pair of straight lines \(x^{2}-y^{2}+2 y-1\) is
1 2 sq. units
2 4 sq. units
3 6 sq. units
4 8 sq. units
Explanation:
(A) : \(\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{y}=1 \Rightarrow \mathrm{x}= \pm(\mathrm{y}-1)\) Bisectors of above line are \(x=0 \& y=1\) So area between \(\mathrm{x}=0, \mathrm{y}=1 \& \mathrm{x}+\mathrm{y}=3\) is shaded Region shown in figure. Area \(=\frac{1}{2} \times 2 \times 2=2\) sq. units
BITSAT-2009
Application of the Integrals
86991
Area bounded by parabola \(y^{2}=x\) and straight line \(2 y=x\) is
1 \(4 / 3\)
2 1
3 \(2 / 3\)
4 \(1 / 3\)
Explanation:
(A) : Parabola \(\mathrm{y}^{2}=\mathrm{x}\) and straight line \(2 \mathrm{y}=\mathrm{x}\) Point of intersection \(\left(\frac{x}{2}\right)^{2}=x\) \(x=0,4\) \(y=0,2\) Area \(\int_{0}^{4} \sqrt{x d x}-\int_{0}^{4}\left(x^{1 / 2}\right) d x\) \(=\frac{2}{3}\left[x^{2 / 3}\right]_{0}^{4}-\left[\frac{x^{2}}{4}\right]_{0}^{4}=\frac{2 \times 8}{3}-4=\frac{4}{3}\)
AMU-2002
Application of the Integrals
86985
What is the area of a loop of the curve \(r=\) \(\operatorname{asin} 3 \theta\) ?
1 \(\frac{\pi \mathrm{a}^{2}}{6}\)
2 \(\frac{\pi \mathrm{a}^{2}}{8}\)
3 \(\frac{\pi \mathrm{a}^{2}}{12}\)
4 \(\frac{\pi \mathrm{a}^{2}}{24}\)
Explanation:
(D) : If curve \(\mathrm{r}=\mathrm{a} \sin 3 \theta\) To trace the curve, we consider the following Table : | $3 \theta=$ | 0 | $\frac{\pi}{2}$ | $\pi$ | $\frac{3 \pi}{2}$ | $2 \pi$ | $\frac{5 \pi}{2}$ | $3 \pi$ | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $\theta=$ | 0 | $\frac{\pi}{6}$ | $\frac{\pi}{3}$ | $\frac{\pi}{2}$ | $\frac{2 \pi}{3}$ | $\frac{5 \pi}{6}$ | $\pi$ | | $\mathrm{r}=$ | 0 | $\mathrm{a}$ | 0 | $-\mathrm{a}$ | 0 | $\mathrm{a}$ | 0 | Thus there is a looping between \(\theta=0 \& \theta=\frac{\pi}{3}\) as \(\mathrm{r}\) varies from \(r=0\) to \(r=0\). Hence, the area of the loop lying in the positive quadrant \(=\frac{1}{2} \int_{0}^{\frac{\pi}{3}} \mathrm{r}^{2} \mathrm{~d} \theta=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin ^{2} \phi \cdot \frac{1}{3} \mathrm{~d} \phi\) [On putting, \(3 \theta=\phi \Rightarrow \mathrm{d} \theta=\frac{1}{3} \mathrm{~d} \phi\) ] \(=\frac{\mathrm{a}^{2}}{6} \int_{0}^{\frac{\pi}{2}} \sin ^{2} \phi \mathrm{d} \phi\) \(=\frac{\mathrm{a}^{2}}{6} \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 \phi}{2} \mathrm{~d} \theta\left[\because \cos 2 \theta=1-2 \sin ^{2} \theta\right]\) \(=\frac{\mathrm{a}^{2}}{12} \cdot\left[\phi+\frac{\sin 2 \phi}{2}\right]_{0}^{\frac{\pi}{2}}=\frac{\mathrm{a}^{2}}{12}\left[\frac{\pi}{2}+\sin \pi\right]=\frac{\pi \mathrm{a}^{2}}{24}\)
86981
The area is square units bounded by the normal at \((1,2)\) to parabola \(y^{2}=4 x, x\)-axis and the curve is given by
1 \(\frac{10}{3}\)
2 \(\frac{7}{3}\)
3 \(\frac{4}{3}\)
4 None of these
Explanation:
(A) : Given, \(\mathrm{y}^{2}=4 \mathrm{x}\) Differentiating with respect to \(\mathrm{x}\), \(2 y \frac{d y}{d x}=4\) \(\frac{d y}{d x}=\frac{2}{y}\) \(\left(\frac{d y}{d x}\right)_{(1,2)}=1\) Equation to normal to the curve at \((1,2)\), \(y-y_{1}=\frac{1}{\left(-\frac{d y}{d x}\right)}\left(x-x_{1}\right)\) \(\Rightarrow(\mathrm{y}-2)=-\frac{1}{1}(\mathrm{x}-1) \Rightarrow \mathrm{y}-2=-\mathrm{x}+1 \Rightarrow \mathrm{x}+\mathrm{y}=3\) Line \(\mathrm{x}+\mathrm{y}=3\) meet the \(\mathrm{x}-\) axis at \(\mathrm{x}=3\) Now required area, \(A=\int_{0}^{1} \sqrt{4 x} \cdot d x+\int_{1}^{3}(3-x) \cdot d x\) \(A=2\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1}+\left[3 x-\frac{x^{2}}{2}\right]_{1}^{3}\) \(\mathrm{A}=\frac{4}{3}+\left[9-\frac{9}{2}-3+\frac{1}{2}\right]\) \(\mathrm{A}=\frac{4}{3}+\left(\frac{9}{2}-\frac{5}{2}\right)=\frac{4}{3}+\frac{4}{2}=\frac{20}{6} \Rightarrow \mathrm{A}=\frac{10}{3}\) square units.
COMEDK-2015
Application of the Integrals
86982
Area of the triangle formed by the line \(x+y=3\) and angle bisectors of the pair of straight lines \(x^{2}-y^{2}+2 y-1\) is
1 2 sq. units
2 4 sq. units
3 6 sq. units
4 8 sq. units
Explanation:
(A) : \(\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{y}=1 \Rightarrow \mathrm{x}= \pm(\mathrm{y}-1)\) Bisectors of above line are \(x=0 \& y=1\) So area between \(\mathrm{x}=0, \mathrm{y}=1 \& \mathrm{x}+\mathrm{y}=3\) is shaded Region shown in figure. Area \(=\frac{1}{2} \times 2 \times 2=2\) sq. units
BITSAT-2009
Application of the Integrals
86991
Area bounded by parabola \(y^{2}=x\) and straight line \(2 y=x\) is
1 \(4 / 3\)
2 1
3 \(2 / 3\)
4 \(1 / 3\)
Explanation:
(A) : Parabola \(\mathrm{y}^{2}=\mathrm{x}\) and straight line \(2 \mathrm{y}=\mathrm{x}\) Point of intersection \(\left(\frac{x}{2}\right)^{2}=x\) \(x=0,4\) \(y=0,2\) Area \(\int_{0}^{4} \sqrt{x d x}-\int_{0}^{4}\left(x^{1 / 2}\right) d x\) \(=\frac{2}{3}\left[x^{2 / 3}\right]_{0}^{4}-\left[\frac{x^{2}}{4}\right]_{0}^{4}=\frac{2 \times 8}{3}-4=\frac{4}{3}\)
AMU-2002
Application of the Integrals
86985
What is the area of a loop of the curve \(r=\) \(\operatorname{asin} 3 \theta\) ?
1 \(\frac{\pi \mathrm{a}^{2}}{6}\)
2 \(\frac{\pi \mathrm{a}^{2}}{8}\)
3 \(\frac{\pi \mathrm{a}^{2}}{12}\)
4 \(\frac{\pi \mathrm{a}^{2}}{24}\)
Explanation:
(D) : If curve \(\mathrm{r}=\mathrm{a} \sin 3 \theta\) To trace the curve, we consider the following Table : | $3 \theta=$ | 0 | $\frac{\pi}{2}$ | $\pi$ | $\frac{3 \pi}{2}$ | $2 \pi$ | $\frac{5 \pi}{2}$ | $3 \pi$ | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $\theta=$ | 0 | $\frac{\pi}{6}$ | $\frac{\pi}{3}$ | $\frac{\pi}{2}$ | $\frac{2 \pi}{3}$ | $\frac{5 \pi}{6}$ | $\pi$ | | $\mathrm{r}=$ | 0 | $\mathrm{a}$ | 0 | $-\mathrm{a}$ | 0 | $\mathrm{a}$ | 0 | Thus there is a looping between \(\theta=0 \& \theta=\frac{\pi}{3}\) as \(\mathrm{r}\) varies from \(r=0\) to \(r=0\). Hence, the area of the loop lying in the positive quadrant \(=\frac{1}{2} \int_{0}^{\frac{\pi}{3}} \mathrm{r}^{2} \mathrm{~d} \theta=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin ^{2} \phi \cdot \frac{1}{3} \mathrm{~d} \phi\) [On putting, \(3 \theta=\phi \Rightarrow \mathrm{d} \theta=\frac{1}{3} \mathrm{~d} \phi\) ] \(=\frac{\mathrm{a}^{2}}{6} \int_{0}^{\frac{\pi}{2}} \sin ^{2} \phi \mathrm{d} \phi\) \(=\frac{\mathrm{a}^{2}}{6} \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 \phi}{2} \mathrm{~d} \theta\left[\because \cos 2 \theta=1-2 \sin ^{2} \theta\right]\) \(=\frac{\mathrm{a}^{2}}{12} \cdot\left[\phi+\frac{\sin 2 \phi}{2}\right]_{0}^{\frac{\pi}{2}}=\frac{\mathrm{a}^{2}}{12}\left[\frac{\pi}{2}+\sin \pi\right]=\frac{\pi \mathrm{a}^{2}}{24}\)
