Learn volume of a right circular cone & volumes of general solids formula for Integration. Practice washer method calculus formula.

- Let S be the solid that lies between \(x=a\) and \(x=b\). If the cross sectional area of \(S\) in plane \(P_x\) through \(x\) and \(\perp\) to x-axis is \(A(x)\) where \(A\) is a continuous function, then volume of \(S\)is given by \(V=\int\limits_a^b\,A(x)\;dx\)

- We cut S into pieces and obtain a plane require that is called cross sectional of S. This cross sectional Area \(A(x)\) will vary as \(x\) increases function \('a'\) to \('b'\).
- Divide S into \(x\) slabs of equal width \(\Delta x\) by using planes \(P_{x_1}, P_{x_2},....,P_{x_n}\) to slice the solid. The \(i^{th}\) slab is like a cylinder with base.
- Area = \(A(x_i^*)\,\big(x_i^*\in(x_{i-1}, \;x_i)\big)\) and height \(\Delta x\)

\(\therefore \) Volume of \(i^x\) slab = \(V(S_i)\approx A(x_i^*)\,\Delta x\)

Adding the volume

\(V\approx\sum\limits_{i=1}^{n}\,A(x_i^*)\,\Delta x\)

as \(n\to\infty\)

\(V=\int\limits_a^b\,A(x)\,dx\)

A \(\dfrac{5\pi}{8}\)

B \(\dfrac{4\pi}{21}\)

C \(6\pi\)

D \(\dfrac{121\pi}{3}\)

A \(\dfrac {64\pi}{15}\)

B \(\dfrac {\pi}{8}\)

C \(\dfrac {32\pi}{5}\)

D \(\dfrac {2\pi}{7}\)

- The volume of a right circular cone whose radius is \(r\) and height is \(h\), is given by

\(V=\dfrac {1}{3}\,\pi r^2\,h\)

A \(\dfrac {10\,\pi}{3}\,m^3\)

B \(\dfrac {20\,\pi}{3}\,m^3\)

C \(5\,\pi\,m^3\)

D \(\dfrac {\pi}{3}\,m^3\)

- The volume of a right circular cone whose radius is \(r\) and height is \(h\), is given by

\(V=\dfrac {1}{3}\,\pi r^2\,h\)

- Consider the line \(y=x×m\) and the region bounded by \(y=x\) and \(x=a\) rotated about x-axis. The figure formed will be a right circular cone.

\(V=\displaystyle\int\limits_0^a\,\pi(mx)^2\;dx=\pi\,m^2\int\limits_0^a\,x^2\;dx\)

\(=\pi\,m^2\;\dfrac {x^3}{3}\Bigg]_0^a\,\)

\(=\pi\,m^2\;\dfrac {a^3}{3}\)

\(=\dfrac {\pi}{3}×a×(m\,a)^2\)

\(=\dfrac {1}{3}\,\pi\,r^2\,h\)

where

\(r=\) radius of cone formed

\(h=\) height of cone

A \(\dfrac {16\,\pi}{3}\)

B \(\dfrac {32\,\pi}{3}\)

C \(\dfrac {4\,\pi}{3}\)

D \(\dfrac {5\,\pi}{3}\)

- We cut S into pieces and obtain a plane region that is called cross-section of S. This cross-sectional Area \(A(x)\) will vary as \(x\) increases from \('a'\) to \('b'\).
- Divide S into \(n\) slabs of equal width \(\Delta x\) by using planes \(P_{x_1}, P_{x_2},....,P_{x_n}\) to slice the solid. The \(i^{th}\) slab is like a cylinder with the base.
- Area = \(A(x_i^*)\,\big(x_i^*\in(x_{i-1}, \;x_i)\big)\) and height \(\Delta x\)

\(\therefore \) Volume of \(i^x\) slab = \(V(S_i)\approx A(x_i^*)\,\Delta x\)

Adding the volume,

\(V\approx\sum\limits_{i=1}^{n}\,A(x_i^*)\,\Delta x\)

as \(n\to\infty\)

\(V=\int\limits_a^b\,A(x)\,dx\)

A \(16\pi\)

B \(2\pi\)

C \(512\pi\)

D \(8\pi\)

- Obtained by rotating a region through a line other than axis.
- Let S be the solid that lies between \(x=a\) and \(x=b\). If the cross sectional area of \(S\) in plane \(P_x\) through \(x\) and \(\perp\) to x-axis is \(A(x)\) where, \(A\) is a continuous function, then volume of \(S\)is given by \(V=\int\limits_a^b\,A(x)\;dx\)

- We cut S into pieces and obtain a plane require that is called cross sectional of S. This cross sectional Area \(A(x)\) will vary as \(x\) increases function \('a'\) to \('b'\).
- Divide S into \(x\) slabs of equal width \(\Delta(x)\) by using planes \(P_{x_1}, P_{x_2},....,P_{x_n}\) to slice the solid. The \(i^{th}\) slab is like a cylinder with base.
- Area = \(A(x_i^*)\,\big(x_i^*\in(x_{i-1}, \;x_i)\big)\) and height \(\Delta x\).

\(\therefore \) Volume of \(i^x\) slab = \(V(S_i)\approx A(x_i^*)\,\Delta x\)

Adding the volume

\(V\approx\sum\limits_{i=1}^{n}\,A(x_i^*)\,\Delta x\)

as \(n\to\infty\)

\(V=\int\limits_a^b\,A(x)\,dx\)

A \(\dfrac {\pi}{28}\)

B \(\dfrac {29\pi}{30}\)

C \(\dfrac {15}{16}\pi\)

D \(\dfrac {22\pi}{3}\)

Consider a region \(R\) enclosed by lines. \(x=0,\;y=x\) and \(y=a\) is rotated about \(x-\) axis.

Lines \(x=0,\;y=x\) and \(y=a\) intersect at the points \((0,\,0),\;(0,\,a)\;\&\;(a,\,a)\) as shown in figure.

Area of cross section is

\(A(R)=\pi\,\left(a^2-x^2\right)\) & volume \(v=\int \limits^a_0\,A(x)\;dx\)

\(V=\int \limits^a_0\,\pi\,\left(a^2-x^2\right)\,dx\)

\(=\pi\,\left[a^2[x]^a_0-\left[\dfrac{x^3}{3}\right]^a_0\right]\)

\(=\pi\,\left(a^3-\dfrac{a^3}{3}\right)\)

\(=\dfrac{2\,a^3}{3}\pi\)

A \(\dfrac{\pi}{2}\)

B \(\pi\)

C \(\dfrac{2\pi}{3}\)

D \(\dfrac{\pi}{3}\)

When solids are obtained by revolving a region about a line, such solids are called as solids of revolution.

- If the cross-section is a washer, inner radius \(r_{in}\) and outer radius \(r_{out}\) can be found from a sketch as shown in the figure.
- Area of the washer can be found by subtracting the area of the inner disk from the area of outer disk.

\(A=\pi\,(\text{Outer radius})^2-\pi\,(\text{Inner radius})^2\)

Further volume can be calculated as

\(V=\int\limits^b_a\,A(y)\,dy\;or\;\int\limits^b_a\,A(x)\,dx\)

A \(\dfrac{16\pi}{3}\)

B \(\dfrac{28\pi}{3}\)

C \(\dfrac{16\pi}{5}\)

D \(\dfrac{11\pi}{3}\)