Integrating vertically giving different answer from intergrating horizontal

Question

Find area in the region: $$ \frac{1}{x} <y < x^{\frac13}$$ , $1 <x<8$

So, I thought of integrating 'vertically' by:

$$ \int y dx = xy - \int x dy =16- \left[ \int_1^2 y^3 + \int_0^1 \frac{1}{y} dy\right] $$

Now, the problem is that $ \int_0^1 \frac{dy}{y}$ part, it seems to blow up. So, this method seems unfeasible, is there are any reason I could 'remove' the blow up and what exactly is the property of the function which causes this to occur?

Note the integral $ \int_0^1 \frac{1}{y} dy$ represents this following area:

Then there is no reason for $\int_0^1 \frac1y dy$ to exist in your steps, and everything should converge. — player3236, Jan 31 '21 at 06:25
You have the integrals wrong. The region that you are highlighting needs to be split into $2$. One is a rectangle plus the area under the curve. $y \ne 0$ for the area under the curve. — Math Lover, Jan 31 '21 at 06:28
the red portion goes all the way to infinity, doesn't it @Math Lover — tryst with freedom, Jan 31 '21 at 06:30
No but $\int_0^1 \frac{1}{y} dy$ does not represent the area that you are claiming. at $x=8, y = 1/8$ so the lower bound of $y$ is $1/8$ and then below that you have a rectangular area. — Math Lover, Jan 31 '21 at 06:31
Ohhhhh @MathLover I don't know why but I kept putting the other functions bound into it. Thank you I understood my mistake now. — tryst with freedom, Jan 31 '21 at 06:34

score 1 · Accepted Answer · answered Jan 31 '21 at 06:31

1

Vertically:

$$\int_{1/8}^1 (8-1/y) \ dy + \int_{1}^2 (8-y^3) \ dy$$

Horizontally:

$$\int_{1}^8 (x^{1/3}-1/x) \ dx$$

answered Jan 31 '21 at 06:31

David P

12,320

Hmm where did you get the eight from? – tryst with freedom Jan 31 '21 at 06:34
If integrating vertically it's a function of $y$. The right-most bound of your region is $x=8$, and the left-most splits between your two functions at $y=1$. – David P Jan 31 '21 at 06:35
Hmm right, but I think this is giving me a different answer from the regular method, the whole rectangles area I got as: $(8-1)( 2 - \frac{1}{8})$ and I subtract $\int_{\frac18} \frac{1}{y} dy + \int_1^2 (y^3) dy$. – tryst with freedom Jan 31 '21 at 06:41
Both integrals are the shaded area you give, and are equal to $45/4-\ln(8)$ – David P Jan 31 '21 at 06:42
8-1 is the length of rectangle along x , and 2-1/8 is width of rectangle along y – tryst with freedom Jan 31 '21 at 06:42
I don;t know why you're concerned with the "rectangle"--it doesn't matter. – David P Jan 31 '21 at 06:43

score 0 · Answer 2 · answered Jan 31 '21 at 08:03

Ok, I basically screwed up the figure, the figure in the question goes from $0 $ to $8$ but we want from $1$ to $8$

The area of rectangle is given as $2 \cdot 8$

The blue area is given as:

$$ \int_1^2 y^3 dy $$

And the purple given as:

$$ \int_{\frac18}^1 \frac{1}{y} dy$$

Putting it together

$$ \text{required area} = 2 \cdot 16 - \int_1^2 y^3 dy - \int_{\frac18}^1 \frac{1}{y} dy$$

And this gives the same as David P's answer

Integrating vertically giving different answer from intergrating horizontal

2 Answers2