Suppose we have $y = x^2$, $y = x^4$, $y = 16$ and $x = 5$
a) integrate with respect to $y$, to calculate area of the region (area enclosed by the above functions)
b) integrate with respect to $x$, to calculate the volume of the region rotated about $x = -1$ (shell method).
I know HOW to integrate, but im having trouble finding WHAT to integrate.For part a, the best i could do was
$$\text{Area} =\text{ Area of rectangle }- \int ^{ 625 }_{ 16 }{ y^{1/4}dy }$$
but i miss a little area ($y = x^2$ between $x = 4$ and $x =5$)
For part a), your approach is on the right track. You will have to subtract the "little area" separately. Alternatively, you could set up your answer as follows: $$ \text{Area} = \left[ \int_{16}^{25} (y^{1/2} - y^{1/4})~dy \right] + \left[ \int_{25}^{625} (5 - y^{1/4})~dy \right] $$
For part b), the radius of each shell will be $x+1$, but the height varies. To the left of the vertical blue line at $x=4$, the height is $x^4 - 16$, while to the right the height is $x^4 - x^2$. Putting everything together, we obtain: $$ \text{Volume} = \left[ \int_{2}^{4} 2\pi (x+1)(x^4 - 16)~dx \right] + \left[ \int_{4}^{5} 2\pi (x+1)(x^4 - x^2)~dx \right] $$