Assuming the area below the line y=0 (i.e. x>1) does NOT count, the area to be rotated is shown in the graph attached.
A. Again, using Pappus's theorem,
Area, A = (2/3)*1*(1-(-1))=4/3 (2/3 of the enclosing rectangle, or you can integrate)
Distance of centroid from axis of rotation, R = (2-0) = 2
Volume = 2 π RA = 2 π 2 * 4/3 = 16 π / 3 (approximately = 16.76 units)
B. By integration, using the washer method
Volume =
= 16 π /3 as before
Answer:
1π
Step-by-step explanation:
suppose the radius of semicircle P is r,
then the radius of semicircle Q = (r+d)/2 ... d≤r
radius of semicircle R = (r-d)/2
area P = 1/2 (r)²π
area Q = 1/2 ((r+d)/2)² π = 1/8 (r² + 2rd + d²)π
area R = 1/2 ((r-d)/2)² π = 1/8 (r² - 2rd + d²)π
shaded area = P-Q-R = 1/2 r²π - 1/4 (r² + d²)π
= ((r² - d²)/4) * π
because there is no constant r value in the question and d value changes with the r change, when the vertical segment length equal the semicircle P radius (r), r=2 and d = 0
therefore the shaded area = ((2² -0²)/4)*π = 1π
Answer:
The value of n is -6
Step-by-step explanation:
- If the function f(x) is translated k units up, then its image is g(x) = f(x) + k
- If the function f(x) is translated k units down, then its image is g(x) = f(x) - k
- The vertex form of the quadratic function is f(x) = a(x - h)² + k, where a is the coefficient of x² and (h, k) is the vertex
∵ k(x) = x²
→ Its graph is a parabola with vertex (0, 0)
∴ The vertex of the prabola which represents it is (0, 0)
∵ The given graph is the graph of p(x)
∵ Its vertex is (0, -6)
∴ h = 0 and k = -6
∵ a = 1
→ Substitute them in the form above
∴ p(x) = 1(x - 0)² + -6
∴ p(x) = x² - 6
→ Substitute x² by k(x)
∴ p(x) = k(x) - 6
∵ p(x) = k(x) + n
→ By comparing the two right sides
∴ n = -6
∴ The value of n is -6
Look at the attached figure for more understanding
The red parabola represents k(x)
The blue parabola represents p(x)
Answer:
Where is the shape?
Step-by-step explanation: