Used shell method with y = sqrt(2x-9) & x axis, revolving around y axis. Volume integral: Π ∫_5^9 (2x-9) dx. Evaluating gives ≈ 62.83 cubic cm.
Imagine the solid as thin slices:
Picture the region between the curve `y = sqrt(2x - 9)`, the x-axis (y = 0), and the vertical line x = 5 being divided into many thin vertical slices.
Each slice is like a cylindrical shell with a hole in the middle, once rotated about the y-axis.
Define the dimensions of each slice:
Let the width of each slice be `dx`.
Let the radius of the base, as a function of x, be `r_1(x)`. This represents the distance from the curve `y = sqrt(2x - 9)` to the y-axis.
Since the hole coincides with the y-axis, the radius of the hole, `r_2(x)`, is always 0.
Express the volume of each slice as a cylindrical shell:**
The volume of a thin cylindrical shell is `πh(r_1^2 - r_2^2)`, where `h` is the height of the shell (which in this case is equal to `dx`).
- In this case, the volume of each slice is `π dx (r_1(x)^2 - 0) = π dx r_1(x)^2`.
Sum the volumes of infinitely many such slices with an infinitely small width using a definite integral:
The definite integral that represents the total volume `V` of the solid is:
V = ∫_a^b π r_1(x)^2 dx
where `a` and `b` are the x-coordinates of the starting and ending points of the curve, respectively.
Apply the specific information to the given graph:
In this case, `r_1(x) = sqrt(2x - 9)`.
The curve starts at `x = 5` and ends at `x = 9` (since the square root becomes undefined before that).
Substitute and evaluate the integral:
V = ∫_5^9 π (sqrt(2x - 9))^2 dx
V = ∫_5^9 π (2x - 9) dx
V = π [(x^2 - 9x)]|_5^9
V = π [(81 - 81) - (25 - 45)]
V = π (-20)
V = -20π (negative volume doesn't make physical sense, so the absolute value is the actual volume)
V = 20π ≈ 62.83 cm^3 (rounded to two decimal places)
Therefore, the definite integral that represents the volume of the solid is:
∫_5^9 π (2x - 9) dx
and its evaluation gives the volume as approximately 62.83 cubic centimeters.