My teacher even knew that the answer should not be negative but it turned out to be negative. The given was y=x^2, y=4x-x^2, revolving about the y-axis. Here are some of the solution presented, I hope someone could help me find where I went wrong while solving. Thanks in advance. My teacher can’t even find the error that I have committed.
Edit: As I did the right - left thingy, the answer was -112/3(pi) cu. units.
V=(pi)[(sqrt(y))^2-(sqrt(4-y)+2)^2)]dy
V=(pi)[2y-4sqrt(4-y)-8]dy
V=(pi)[y^2+8/3(4-y)^3/2-8y] limit from 0 to 4 <——already integrated
Applying the limits
V=(pi)[16+8/3(0-8)]
V=-112/3(pi) cu. units. Is there anything done wrong? I hope someone can spot what is wrong. I really need the correct answer utgently. Thanks for those who can help.
Hint: if you plan on using the $y$-axis as your axis of rotation for the washer method, then you are going to need to integrate with respect to $y$, and you need to solve each of the equations in terms of $x$: $$y = x^2 \implies \pm\sqrt y = x \implies \sqrt y = x \checkmark \qquad\text{(Since we need the right-half)}$$ However, the other equation is not so straightforward. You will need to complete the square to help eliminate one of the $x$'s: $$\begin{align}y &= 4x-x^2\\ &= -(x^2-4x + 4 - 4) \\ &= -(x-2)^2 + 4 \\ \implies y- 4 &= -(x-2)^2\\ \pm\sqrt{4-y} &= x-2 \\ \implies \pm\sqrt{4-y}+2 &= x\\ \implies -\sqrt{4-y}+2 &= x. \qquad\text{(Since we need the left-half)}\end{align}$$
Notice how $\sqrt y = x$ is the rightmost graph and $-\sqrt{4-y}+2 = x$ is the leftmost. You will need to do right - left in your radius function when computing the integral.