Volumes (Cross Sections Method)

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Volumes - Cross Sections Method
The volume of a solid with parallel cross sections that are perpendicular to the base is given by
V=∫x=ax=bA(x) dx  or  ∫y=cy=dA(y) dy\displaystyle V=\int_{x=a}^{x=b}A\left(x\right)\ dx\ \ \text{or}\ \ \int_{y=c}^{y=d}A\left(y\right)\ dyV=∫x=ax=b​A(x) dx  or  ∫y=cy=d​A(y) dy
where A(x)  or  A(y)A\left(x\right)\ \ \text{or}\ \ A\left(y\right)A(x)  or  A(y) are functions representing the areas of these cross sections.

Example
Write an integral that represents the volume of a cube of side length LLL

Since the cross sections that are perpendicular to the base of the cube are all squares that have the same size, the cross sectional area is just L2L^2L2
Let's suppose that one end of this cube starts at x=0x=0x=0 and goes to x=Lx=Lx=L. Then, using our formula, we know that the volume is 
V=∫0LL2dx=[L2x]0L=L3\displaystyle V=\int_0^LL^2dx=\left[L^2x\right]_0^L=L^3V=∫0L​L2dx=[L2x]0L​=L3

Watch Out!
They will typically ask you to find the volume of a solid that does NOT have equally-sized cross sections, instead, the area of the cross section will usually be based on some function.

Practice: Volume Using Cross Sections
Suppose that a solid has a circular base of radius aaa. The parallel cross sections that are perpendicular to the base are isosceles triangles where the length of the two identical sides are twice the length of the base of the triangle. Find the integral representing the volume of this solid.

V=∫−aa(a2−x2)dx\displaystyle V=\int_{-a}^a(a^2-x^2)dxV=∫−aa​(a2−x2)dx

V=∫−aaa2−x22dx\displaystyle V=\int_{-a}^a\frac{a^2-x^2}{2}dxV=∫−aa​2a2−x2​dx

V=∫−aa15(a2−x2)dx\displaystyle V=\int_{-a}^a\sqrt{15}\left(a^2-x^2\right)dxV=∫−aa​15​(a2−x2)dx

V=∫−aay2 dy\displaystyle V=\int_{-a}^ay^2\ dyV=∫−aa​y2 dy

V=∫−aa3y2 dy\displaystyle V=\int_{-a}^a\sqrt{3}y^2\ dyV=∫−aa​3​y2 dy

I don't know

Practice: Volumes Using Cross Sections
Find the volume of a cone whose base is a circle with radius RRR and whose height is hhh.

Hint: R and h are constants, the only variable you should use is xxx.

πR2h22\displaystyle \frac{\pi R^2h^2}{2}2πR2h2​

πR2h\displaystyle \pi R^2hπR2h

πR2h3\displaystyle \frac{\pi R^2h}{3}3πR2h​

2πR2h3\displaystyle \frac{2\pi R^2h}{3}32πR2h​

3πR2h2\displaystyle \frac{3\pi R^2h}{2}23πR2h​

I don't know

Wize University Calculus 2 Textbook > Applications of Integrals

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