I've come across this problem on several calculus tutorials but can't find any solutions for it. Can someone please explain how to figure these questions out?
For question #1 I found the link below that helped me figure it out:
Volume of a Torus: the Washer Method
ANSWER for question #1:
But I can't find any help on how to solve problems 2, 3, and 4 based on the instructions.
On
Item 2
The derivation of the surface area formula actually exists already on this site, here. You can also see it done in this YouTube video.
Item 3
We want to find the minimum surface area for the fixed volume of $90 \pi^2$. We know that the volume $V$ and the surface area $S$ is given by the formulas $$V=2 \pi^2 a b^2 = 90 \pi^2 \tag {1}$$
$$S = 4 \pi^2 ab \tag {2}$$ From (1) we can find $$a = \frac {45}{b^2} \tag {3}$$ Inserting this in (2) gives $$S=\frac {180 \pi^2}{b} \tag{4}$$ We want $S$ as small as possible so $b$ must be as large as possible. But for a torus $b \le a$, otherwise it overlaps itself. So $b = a$, which using (3) gives $a = b = \sqrt[3]{45}$, and the minimum surface area is then $$S_{min} = 4 \pi^2 45^{\frac 2 3 } \approx 50.6 \pi^2$$
Item 4
Equation (4) shows there is no maximum surface area as $S \to \infty$ as $b \to 0$.
On
Let's assume the formulae $V= 2\pi^2ab^2,$ $A = 4\pi^2 ab$ are known. Then, as noticed by others, $A = (2V)/b .$ Thus if we assume $V$ is fixed at some $V_0,$ we are trying to minimize $(2V_0)/b.$ That is the same as maximizing $b$ given the constraints of the situation.
One of the constraints is $0\le b\le a.$ The other one is
$$\tag 1 b= \left ( \frac{V_0}{2\pi^2a}\right)^{1/2}.$$
Looking at the $a$-$b$ plane, we want to consider the intersection in the first quadrant of the curve $b = V_0/(2\pi^2a)^{1/2}$ with the region $0\le b\le a.$ The intersection point with the largest $b$ occurs where the curve intesects the line $b=a.$ Thus we only need solve $V_0 = 2\pi^2a^3,$ which gives $a=b= (V_0/(2\pi^2))^{1/3}$ as yielding the minimum surface area.
If we take $b$ tiny and $a$ large so as to have $V_0= 2\pi^2ab^2,$ it is clear that $A = (2V_0)/b$ can be made arbitrarily large. Thus there is no maximum surface area for a fixed volume.
As for the surface area formula $A = 4\pi^2 ab,$ we can obtain this by noticing the upper half of the torus is the solid swept out by revolving the upper half of the circle $(x-a)^2 + y^2= b^2$ $360$ degrees about the $y$-axis. Solving for $y$ in the upper half circle gives
$$y = \sqrt { b^2 - (x-a)^2}.$$
The desired surface area is thus
$$2\cdot 2\pi \int_{a-b}^{a+b}x\cdot \sqrt {1+ (dy/dx)^2}\, dx.$$
This works out nicely. Ask if you have questions on it.
I will let you solve for the volume and surface area any way you choose. Here I'll just give the results using Pappus's centroid theorems. The solutions are
$$ V=2\pi R A=2\pi^2 Rr^2\\ S=2\pi R C=4\pi^2 Rr $$
where $R$ is the centroid of the revolving circle, and $A=\pi r^2$ and $C=2\pi r$ are its area and circumference, respectively. Here, $R$ and $r$ correspond to $a$ and $b$ in the problem.
Then we find that we can express
$$S=\frac{2V}{r}$$
If $V$ is fixed, then the maximum surface area coincides with smallest radius. In this problem, the smallest radius is equal to $R$, i.e., the doughnut with no hole has the largest surface area. As the hole increases in size (for a fixed volume) the surface area necessarily decreases.