A bagel is obtained by revolving a circle of radius $ \ \frac{3}{4} \ $ around the y-axis centered at point $ \ (1,0) \ $. The equation of the circle is $ \ (x-1)^2 + y^2 = (3/4)^2 \ $.
Use calculus to set up an integral for the volume of the bagel using washer method.
Answer:
Let $ \ V=volume \ $ , then using washer method
$V= \int_{-\frac{3}{4}}^{\frac{3}{4}} \pi (R^2-r^2) dy \\ $
where $ \ R=outer \ \ radius \ $ and $ \ r=inner \ \ radius \ $
But what would be inner radius and outer radius?
On
$$\int_{-3/4}^{3/4} \pi x^2 dy$$
That's a setup for a disk method, not a washer method. You need to explicitly indicate the "hole", not just a single radius.
Since you don't have enough symbols in the problem statement to write the formulas you need, you will have to use some words to define new symbols.
$$\pi \int_{-3/4}^{3/4} \left( 1\pm \sqrt{\left(\frac34\right)^2-y^2} \right) dy$$
What does that even mean? Does it mean there are two possible answers, one in the $+$ case and one in the $-$ case?
Also, it looks like you forgot about the $2$ in $x^2$ when you substituted $x.$
Let $a=1$ be the radius of the "soul" of the bagel, and $b={3\over4}$ be the radius of the meridian circle. Your formula for the volume is basically correct, but you should clearly exhibit that the outer and inner radii of the washers depend on $y$: $$V=\pi\int_{-b}^b \bigl(R^2(y)-r^2(y)\bigr)\>dy\ .$$ Now from the figure we can deduce that $$R(y)=a+\sqrt{b^2-y^2},\quad r(y)=a-\sqrt{b^2-y^2}\qquad(-b\leq y\leq b)\ .$$ It follows that $$R^2(y)-r^2(y)=4a\sqrt{b^2-y^2}\ ,$$ so that $$V=4a\pi\int_{-b}^b\sqrt{b^2-y^2}\>dy\ ,$$ which I may leave to you.