Simple Definite Integral: Find the Area Bounded by a Curve

[Mentallic]

Homework Statement

Find the area bounded by the curve $$y=x^2-2$$, the y-axis, y=0 and y=1

I got an answer, but when I checked it using graphmatica, it was wrong (also the result is logically too large). I cannot see where I went wrong though, so could someone please help me spot the mistake.

Homework Equations

$$\int_{a}^{b}f(y)dy = \left [ F(y) \right ]_{a}^{b}$$

$$\int_{a}^{b}(ax+b)^ndx=\left [ \frac{(ax+b)^{n+1}}{a(n+1)}\right ]_a^b$$

The Attempt at a Solution

$$A=\int_{0}^{1}xdy$$

where $$x=\pm \sqrt{y+2}$$

thus,
$$A=\int_{0}^{1}\pm \sqrt{y+2}.dy$$

$$A=2\int_{0}^{1}\sqrt{y+2}.dy$$

$$=2 \left[ \frac{(y+2)^{3/2}}{3/2} \right]_0^1$$

$$=2\left( \frac{3^{3/2}}{3/2} \right)$$

$$=4\sqrt{3}.u^2$$

 

[lanedance]

hmm... i think it could be the factor of 2, the question says from y axis, so try only taking +sqrt

 

[Mentallic]

Uh no actually, there is something else going on here.

This is why:

Using graphmatica, I found $$\int_0^1 \sqrt{x+2}.dx \approx 1.5785$$

I changed the function to integrate because the program seems to only be able to integrate with respect to the x-axis. The function is still equivalent to the original question though (for positive values of y).

So if I take the area of $$\int_0^1 \sqrt{y+2}.dy$$ in the positive quadrant only, then I get $$2\sqrt{3} \approx 3.464$$ which is far off the value given from graphmatica (and this answer doesn't seem correct either, from an elementary view of the approx area).

The problem is somewhere else, and I can't seem to spot it...

 

[Mentallic]

Ahh I found where I went wrong:

$$\left[ \frac{(y+2)^{3/2}}{3/2} \right]^1_0 \neq 2\sqrt{3}$$

I assumed substituting 0 into the primitive returned 0. I'll be sure not to make that mistake again..

 

[HallsofIvy]

Homework Statement

Find the area bounded by the curve $$y=x^2-2$$, the y-axis, y=0 and y=1

I got an answer, but when I checked it using graphmatica, it was wrong (also the result is logically too large). I cannot see where I went wrong though, so could someone please help me spot the mistake.

Homework Equations

$$\int_{a}^{b}f(y)dy = \left [ F(y) \right ]_{a}^{b}$$

$$\int_{a}^{b}(ax+b)^ndx=\left [ \frac{(ax+b)^{n+1}}{a(n+1)}\right ]_a^b$$

The Attempt at a Solution

$$A=\int_{0}^{1}xdy$$

where $$x=\pm \sqrt{y+2}$$

That's makes no sense. You must integrate a function and "$\pm\sqrt{y+2}$" is not a function. The x distance from one side of the graph to the other is $2\sqrt{y+2}$. Further, [tex]\int_0^1 xdx[/itex] is NOT the same as $$\int_0^1\sqrt{y+ 2}dy$$. If you let $$x= \sqrt{y+2}= (y+ 2)^{1/2}$$, then $$dx= (1/2)(y+2)^{-1/2}dy= 1/(2x)dy$$ so $$dy= 2xdx$$. Also when y= 0, $x= \sqrt{2}$ and when y= 1, $x= \sqrt{3}$.<br /> $$\int_0^1 \sqrt{y+2}dy= 2\int_{\sqrt{2}}^{\sqrt{3}} x^2dx= \frac{2}{3}\left(\sqrt{27}- \sqrt{8}\right)$$ so your area is $(4/3)\left(\sqrt{8}- \sqrt{2}\right)$.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> thus, <br /> $$A=\int_{0}^{1}\pm \sqrt{y+2}.dy$$<br /> <br /> $$A=2\int_{0}^{1}\sqrt{y+2}.dy$$<br /> <br /> $$=2 \left[ \frac{(y+2)^{3/2}}{3/2} \right]_0^1$$<br /> <br /> $$=2\left( \frac{3^{3/2}}{3/2} \right)$$<br /> <br /> $$=4\sqrt{3}.u^2$$ </div> </div> </blockquote>[/tex]

 

[Mentallic]

That's makes no sense. You must integrate a function and "$\pm\sqrt{y+2}$" is not a function.

How so? if $y=x^2-2$ is a function, then if x is made the subject, why isn't it a function anymore?

The x distance from one side of the graph to the other is $2\sqrt{y+2}$. Further, [tex]\int_0^1 xdx[/itex] is NOT the same as $$\int_0^1\sqrt{y+ 2}dy$$. If you let $$x= \sqrt{y+2}= (y+ 2)^{1/2}$$, then $$dx= (1/2)(y+2)^{-1/2}dy= 1/(2x)dy$$ so $$dy= 2xdx$$. Also when y= 0, $x= \sqrt{2}$ and when y= 1, $x= \sqrt{3}$.[/tex]

[tex] <br /> Yes I did this, but you could argue I made it in a less conventional (cheating) way <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" /><br /> <br /> <blockquote data-attributes="" data-quote="HallsofIvy" data-source="post: 2093184" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> HallsofIvy said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> $$\int_0^1 \sqrt{y+2}dy= 2\int_{\sqrt{2}}^{\sqrt{3}} x^2dx= \frac{2}{3}\left(\sqrt{27}- \sqrt{8}\right)$$ so your area is $(4/3)\left(\sqrt{8}- \sqrt{2}\right)$. </div> </div> </blockquote>The area you found was correct but you simplified it incorrectly.<br /> <br /> $$\frac{2}{3}\left(\sqrt{27}- \sqrt{8}\right) \neq (4/3)\left(\sqrt{8}- \sqrt{2}\right)$$<br /> <br /> The answer $A=\frac{4}{3}(3\sqrt{3}-2\sqrt{2}) \approx 3.157$ is also what Graphmatica gave me. Now the question as to whether the area is actually half of this or not. The question says it is also bounded by the y-axis, but $y=x^2-2$ has x values of both negative and positive between $0<y<1$ to the y-axis...[/tex]