Help please?

I came across this question in a book.

Let \(x\) and \(a\) stand for distance. Is \[\int\frac{dx}{\sqrt{a^{2}-x^{2}}}=\frac{1}{a}\sin^{-1}\frac{a}{x}\] dimensionally correct?

And this is a solution I found.

Dimension of the left side =∫dxa2−x2=∫LL2−L2=[L0]=\int\frac{dx}{\sqrt{a^{2}-x^{2}}}=\int\frac{L}{\sqrt{L^{2}-L^{2}}}=[L^{0}]

Dimension of the right side =1asin⁡−1ax=[L−1]=\frac{1}{a}\sin^{-1}\frac{a}{x}=[L^{-1}]

So, the dimension of ∫dxa2−x2≠1asin⁡−1ax\int\frac{dx}{\sqrt{a^{2}-x^{2}}}\neq\frac{1}{a}\sin^{-1}\frac{a}{x}

So, the equation is dimensionally incorrect.

Could someone explain me how ∫LL2−L2=[L0]\int\frac{L}{\sqrt{L^{2}-L^{2}}}=[L^{0}]? Shouldn't the denominator be zero or something?

#Dimensions #HelpMe!

Note by Omkar Kulkarni
6 years ago

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1 vote

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Comments

Remember we're not talking about numbers anymore. We're talking about dimensions. Rules of algebra do not nessecarily apply here. So [L2]−[L2] [L^2] - [L^2] is not nessecarily 0 0 . Both x2 x^2 and a2 a^2 have a dimesnion of area. So when we add them or subtact them, the resulting quantity will also have the same dimension of area. (as long as the sum/differnce is not 0 ).

Look at it this way. If two sqaures have an area of 9m2 9 m^2 and 4m2 4 m^2 , what is the difference of the areas? 5m2 5m^2 . The equation here is 9m2−4m2=5m2 9m^2 - 4m^2 = 5m^2 . Or, if you look at this dimensionally, [L2]−[L2]=[L2] [L^2] - [L^2] = [L^2]

Siddhartha Srivastava - 6 years ago

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Oh okay! I overlooked that. Thanks! :D

Omkar Kulkarni - 6 years ago

i think the answer to your integral is wrong ,the given integral is :-

∫dxa2−x2=arcsin⁡xa+k\int { \frac { dx }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } } =\arcsin { \frac { x }{ a } } +k

now recheck the dimensions.

u can recheck for yourself ,just substitute x=x=asin⁡θ a\sin { \theta }

Soumya Dubey - 6 years ago

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Yeah, there's a similar question which has sin⁡−1[xa+1]\sin^{-1}\left[\frac{x}{a}+1\right]. But this integral is meant to be wrong. Thanks though! But why can you substitute x=asin⁡θx=a\sin\theta?

Omkar Kulkarni - 6 years ago
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