Answer:
32π²
Step-by-step explanation:
To solve this, we can use the washer method.
We can start by dividing the donut into horizontal sections. Each horizontal section is a hollow cylinder, with the inner cylinder's radius being the difference between x=4 and x²+y²=4, which can also be written as x = √(4-y²) (we do not need the negative x values as the inner cylinder does not encompass any part of the circle). This is (4-√(4-y²)) .
Similarly, the outer radius will be equal to the difference between 4 and the other side of the circle. This encompasses the circle and the extra area in between, so x will be negative here, making it (4-(-√(4-y²)) = 4+√(4-y²).
The area of the cylinder for a given Δy can be represented by
(4+√(4-y²))²πΔy - (4-√(4-y²))²πΔy. This space represents a cylinder in the donut created with the height being Δy. Simplifying the equation, we get
(4+√(4-y²))²πΔy - (4-√(4-y²))²πΔy = πΔy((4+√(4-y²))²- (4-√(4-y²))²)
(4+√(4-y²))²- (4-√(4-y²))² = 16+8√(4-y²) + 4-y² - (16-8√(4-y²)+4-y²)
= 16√(4-y²)
(4+√(4-y²))²πΔy - (4-√(4-y²))²πΔy = 16√(4-y²)πΔy
The donut ranges from all y values from -2 to 2, and to add all the areas of the cylinders up between these y values, we can make this an integral,
[tex]\int\limits^2_{-2} {16\pi \sqrt{4-y^2} \, dy\\[/tex]
perform u substitution, make y = 2sin(u). If y=2sin(u), then y/2 = sin(u) and arcsin(y/2) = u
[tex]\int\limits^2_{-2} {16\pi* 2cos(u) \sqrt{4-4sin^2(u)} \, du\\[/tex]
take the 32π out of the integral
[tex]32\pi\int\limits^2_{-2} {cos(u) \sqrt{4-4sin^2(u)} \, du\\\\= 32\pi\int\limits^2_{-2} {cos(u) *2cos(u)} \, du\\\\\\= 32\pi\int\limits^2_{-2} {2cos&^2(u)} \, du\\\\\\\\\\[/tex]
take the 2 out of the integral
[tex]64\pi\int\limits^2_{-2} {cos&^2(u)} \, du\\[/tex]
One reduction formula we can use here is that
[tex]\int\limits {cos^nx} \, dx = \frac{1}{n} cos^{n-1}(x) sin(x) +\frac{n-1}{n} \int\limits {cos^{n-2} (x)} \, dx[/tex]
Applying that here, we get
[tex]\int\limits {cos^2u} \, du = 0.5cos(u)sin(u) + 0.5 \int\limits {1} \, du\\= 0.5cos(arcsin(y/2))sin(arcsin(y/2)) + 0.5(arcsin(y/2))[/tex]
take that cos(arcsin(x)) = √(1-x²) and sin(arcsin(x)) = x and we get
[tex]\int\limits {cos^2u} \, du = 0.5\sqrt{1-y^2/4}*(y/2) + 0.5arcsin(y/2)\\= 0.25y\sqrt{1-y^2/4} + 0.5arcsin(y/2)\\[/tex]
multiply this back with the 64π to get
[tex]16\pi y\sqrt{1-y^2/4} + 32\pi arcsin(y/2)[/tex]
as our integral. Applying this to the bounds of [-2, 2], we get
16π(2)√(1-4/4) + 32πarcsin(1) - (16π(-2)√(1-4/4) + 32πarcsin(-1))
= 0 + 32π(π/2) - 32π(-π/2)
= 16π²+16π²
= 32π²
as our answer
