I've been met with this problem:
Find the volume of the solid generated by revolving the shaded region about the y-axis (using the slicing method).
I've provided an image of the graph below.
My problem: Even though I've handled the tricky part to the problem, I don't understand where I could've gone wrong. I know that within the problem, you must turn $tan^2u$ into $sec^2u-1$ as per the Pythagorean identity and then integrate.
My work: The setup $$\pi\int_0^1[2tan(\frac{\pi}4y)]^2dy$$ $$\pi\int_0^14tan^2(\frac{\pi}4y)dy$$ $$4\pi\int_0^1tan^2(\frac{\pi}4y)dy$$ u-substitution - Let $u = \frac{\pi}4y$ $$du=\frac{\pi}4dy$$ $$dy=\frac{4(du)}\pi$$ Now replace dy with what dy is equal to $$4\pi\int_0^1tan^2(u)*\frac{4(du)}\pi$$ $\pi$ cancels out and 4 gets multiplied by 4 $$16\int_0^1tan^2(u)*du$$ Use the Pythagorean identity $$16\int_0^1sec^2(u)-1*du$$ Integrate $$16[sin(u)-u]_0^1$$ Substitute u back in $$16[sin(\frac{\pi}4y)-\frac{\pi}4y]_0^1$$
I then plug in 1 for y (I don't bother with 0 since all terms have a variable y) and I get $16sin(\frac{\pi}4)-4\pi$. I then simplify for my final answer as $$4(4sin(\frac{\pi}4)-\pi)$$
However, mymathlab's answer is only slightly different from mine $$4(4-\pi)$$
I don't know where I went wrong, but any help is greatly appreciated!
EDIT: Accidentally included a y inside my final answer (copying error)
As I mentioned in a comment, the antiderivative of $\sec^2x$ is $\tan x +C$. However, there is a part of your solution that needs to be looked at.
When performing substitution, always remember that you have to consider the bounds of your definite integral. As such, the substitution line in your solution should read
$$4\pi\int_0^1\tan^2(\frac{\pi}{4}y)\ dy=4\pi\ (\frac{4}{\pi})\int_0^\frac{\pi}{4}\tan^2 u\ du.$$
Then, when integrating, it is unnecessary to substitute back $u=\frac{\pi}{4}y$. Simply:
\begin{align} 4\pi\ (\frac{4}{\pi})\int_0^\frac{\pi}{4}\tan^2 u\ du&=16\int_0^{\frac{\pi}{4}}\sec^2u - 1\ du\\ &=16\bigg[\tan u-u\bigg]_0^{\frac{\pi}{4}}\\ &=16\bigg((\tan\frac{\pi}{4}-\frac{\pi}{4})-(\tan 0 - 0)\bigg)\\ &=16(1-\frac{\pi}{4})\\ &=4(4-\pi) \end{align}