Applications of definite integral

1103068303

Level: 
B
Which of the following formulas will not give the volume of the solid created by the rotation of the red area about the \( x \)-axis (see the picture)?
\( \pi\cdot\int\limits_{\frac{\pi}4}^{\frac{3\pi}4}\sin x\,\mathrm{d}x \)
\( \pi\cdot\int\limits_{\frac{\pi}4}^{\frac{3\pi}4}\sin^2⁡x\,\mathrm{d}x \)
\( 2\pi\cdot\int\limits_{\frac{\pi}2}^{\frac{3\pi}4}\sin^2x\,\mathrm{d}x \)
\( \pi\cdot\int\limits_{\frac{9\pi}4}^{\frac{11\pi}4}\sin^2x\,\mathrm{d}x \)

1103118704

Level: 
B
Which of the given equations defines the line that together with \( x=0 \) and \( x \)-axis bounds the right triangle, if by rotation of this triangle about the \( x \)-axis the cone of height \( 10 \) is obtained as indicated in the picture?
\( y=-0.4x+4 \)
\( y=-2.5x+10 \)
\( y=4x+10 \)
\( y=10x+4 \)

2010012605

Level: 
B
The function \(f(x) = \frac12 x +2\) is graphed in the picture. Consider the region between the graph of the function \(f\), the \(x\)-axis and the lines \(x = -2\) and \(x = 1\). Find the volume of the solid of revolution obtained by revolving this region about \(x\)-axis.
\(\frac{39} {4} \pi \)
\(\frac{55} {4} \pi \)
\(3\pi \)
\(\frac{10} {3} \pi \)

2010012606

Level: 
B
Part of the graph of the function \(f(x) = \frac{1} {x^2}\) is shown in the picture. Consider the region bounded by \(x\)-axis, graph of \(f\) and lines \(x = 1\) and \(x = 2\). Find the volume of the solid of revolution obtained by revolving this region about \(x\)-axis.
\(\frac{7} {24} \pi \)
\(\frac{\pi} {2}\)
\(\frac{9} {24} \pi \)
\(\frac{7} {8} \pi \)

9000100001

Level: 
B
The function \(f(x) = 3 - 2x\) is graphed in the picture. Consider the region between the graph of the function on the interval \([ 0;\, 1.5] \) and the axes. Determine the solid of revolution obtained by revolving this region about \(y\)-axis
A cone with the base of radius \(1.5\).
A cone with the base of radius \(3\).
A pyramid of the height \(1.5\).
A pyramid of the height \(3\).

9000100002

Level: 
B
The function \(f(x) = 3 - 2x\) is graphed in the picture. Consider the region between the graph of the function \(f\), the \(x\)-axis and the lines \(x = 1\) and \(x = -1\). Find the volume of the solid of revolution obtained by revolving this region about \(x\)-axis.
\(\frac{62} {3} \pi \)
\(6\pi \)
\(12\pi \)
\(\frac{8} {3}\pi \)

9000100003

Level: 
B
The function \(f(x) = x^{2} + 2\) is graphed in the picture. Consider the region between the graph of the function on the interval \([ 0;\, 1] \), both axes and the line \(x = 1\). Find the formula for the volume of the solid of revolution obtained by revolving this region about \(y\)-axis.
\(V =\pi \int _{ 0}^{3}1\, \mathrm{d}y -\pi \int _{2}^{3}(\sqrt{y - 2})^{2}\, \mathrm{d}y\)
\(V =\pi \int _{ 0}^{3}(\sqrt{y - 2})^{2}\, \mathrm{d}y\)
\(V =\pi \int _{ 2}^{3}(\sqrt{y - 2})^{2}\, \mathrm{d}y -\pi \int _{0}^{3}1\, \mathrm{d}y\)
\(V =\pi \int _{ 2}^{3}(\sqrt{y - 2})^{2}\, \mathrm{d}y\)