Let the region bounded by x^2 + y^2 = 9 be the base of a solid. Find the volume if cross sections taken perpendicular to the base are isosceles right triangles.
(a) 30
(b) 32
(c) 34
(d) 36
(e) 38
32
To find the volume of the solid, we need to determine the height of each isosceles right triangle.
Since the base of the solid is given by the equation x^2 + y^2 = 9, we can rewrite this equation as y = sqrt(9 - x^2).
Now, let's consider a cross section of the solid taken at a specific x-value. This cross section will be an isosceles right triangle, so two sides of the triangle will have equal lengths. Let's label the length of these equal sides as s.
Since the cross section is an isosceles right triangle, the height of the triangle can be found using the Pythagorean theorem: h = sqrt(s^2 - (s/2)^2).
To find the total volume of the solid, we integrate the area of each cross section over the range of x-values that define the base.
The range of x-values that define the base can be determined by solving the equation x^2 + y^2 = 9 for x:
-3 ≤ x ≤ 3.
Now we can set up the integral to find the volume:
V = ∫[from -3 to 3] A(x) dx,
where A(x) is the area of the cross section at a specific x-value.
To find A(x), we need to multiply the base length (s) by the height (h) of the triangle:
A(x) = s * h = s * sqrt(s^2 - (s/2)^2).
Now, let's substitute s = sqrt(9 - x^2) into the equation for A(x):
A(x) = sqrt(9 - x^2) * sqrt((9 - x^2)^2 - ((9 - x^2)/2)^2).
Now we can proceed to evaluate the integral.
To find the volume of the solid with cross sections that are isosceles right triangles, we need to integrate the area of each cross section over the region bounded by the base.
Here's how we can approach the problem step by step:
1. First, let's visualize the base of the solid by graphing the equation x^2 + y^2 = 9. This equation represents a circle with radius 3 centered at the origin.
2. Now, let's consider a cross section taken perpendicular to the base. Since each cross section is an isosceles right triangle, we can imagine a triangle inscribed within the circle.
3. Notice that the hypotenuse of the inscribed triangle is the diameter of the circle, which is 2 times the radius, or 6. The legs of the triangle are equal in length and form a right angle.
4. To find the length of each leg of the triangle, we can use the Pythagorean theorem. Let one leg of the triangle be x, then the other leg would be 6 - x.
5. Applying the Pythagorean theorem: x^2 + (6 - x)^2 = 3^2
Simplifying the equation: x^2 + 36 - 12x + x^2 = 9
Combining like terms: 2x^2 - 12x + 27 = 0
6. Solving the quadratic equation, we find two possible values for x: x = 3 and x = 4.5.
7. Since each cross section is an isosceles right triangle, the length of each leg will be equal to x. Therefore, the length of each leg can be 3 or 4.5.
8. Now, let's calculate the area of a cross section. The area of an isosceles right triangle is given by (1/2) * leg^2. So, the area of each cross section is (1/2) * x^2.
9. We need to find the area of each cross section and integrate it over the region bounded by the base. Since the base is a circle with radius 3, we need to integrate the area over the range of -3 to 3.
10. Calculating the integral: ∫(-3 to 3) [(1/2) * x^2] dx
Using the power rule of integration: (1/2) * ∫(-3 to 3) x^2 dx
Applying the power rule: (1/2) * [x^3/3] evaluated from -3 to 3
11. Evaluating the integral: (1/2) * [(3^3/3) - ((-3)^3/3)]
Simplifying: (1/2) * [(27/3) -((-27)/3)]
Calculating: (1/2) * [(9) - (-9)]
Simplifying: (1/2) * [18] = 9
12. Therefore, the volume of the solid with isosceles right triangle cross sections is 9 cubic units.
Since none of the given options match the calculated volume, it seems there might be an error in the problem statement or options.