MHB Is My Double Integral Solution Correct?

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I must evaluate the following double integral over the region R. I do not understand the limits of integration given the following equations.

Let S S = double integral symbol

S S x dx dy

Limits of integration for x:

From (4y/3) to sqrt{25 - y^2}.

Note: Why is the variable y in the radicand? Shouldn't it be sqrt{25 - x^2}?

Limits of integration for y:

From 0 to 3.

I was able to reduced the double integral to a single integral S.

S [(25 - y^2)y]/2 - [(8y^2)/9]y dy from 0 to 3.

My answer is 225/8.

The textbook answer is simply 25.

Is the textbook answer correct? If so, why?
 
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This is how I would evaluate the given integral:

$$I=\int_0^3 \int_{\frac{4}{3}y}^{\sqrt{25-y^2}} x\,dx\,dy$$

$$I=\frac{1}{2}\int_0^3 (25-y^2)-\left(\frac{16}{9}y^2\right)\,dy$$

$$I=\frac{25}{18}\int_0^3 9-y^2\,dy$$

$$I=\frac{25}{54}\left[27y-y^3\right]_0^3=\frac{25}{54}\cdot54=25$$
 
MarkFL said:
This is how I would evaluate the given integral:

$$I=\int_0^3 \int_{\frac{4}{3}y}^{\sqrt{25-y^2}} x\,dx\,dy$$

$$I=\frac{1}{2}\int_0^3 (25-y^2)-\left(\frac{16}{9}y^2\right)\,dy$$

$$I=\frac{25}{18}\int_0^3 9-y^2\,dy$$

$$I=\frac{25}{54}\left[27y-y^3\right]_0^3=\frac{25}{54}\cdot54=25$$

Can you show me how to integrate this same function over dydx?
 
Harpazo said:
Can you show me how to integrate this same function over dydx?

Sure, so what we need to do here is examine the region $D$ over which we are integrating...which can be expressed by the inequalities:

$$0\le y\le3$$

$$\frac{4}{3}y\le x\le\sqrt{25-y^2}$$

Now, in order to reverse the order in integration, we will have to break $D$ up into two regions as follows:

View attachment 6548

$$I=\int_0^4 x\int_0^{\frac{3}{4}x}\,dy\,dx+\int_4^5 x\int_0^{\sqrt{25-x^2}}\,dy\,dx$$

$$I=\frac{3}{4}\int_0^4 x^2\,dx+\int_4^5 x\sqrt{25-x^2}\,dx$$

$$I=\frac{1}{4}\left[x^3\right]_0^4+\frac{1}{2}\int_0^9 u^{\frac{1}{2}}\,du$$

$$I=16+\frac{1}{3}9^{\frac{3}{2}}=16+9=25$$
 

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MarkFL said:
Sure, so what we need to do here is examine the region $D$ over which we are integrating...which can be expressed by the inequalities:

$$0\le y\le3$$

$$\frac{4}{3}y\le x\le\sqrt{25-y^2}$$

Now, in order to reverse the order in integration, we will have to break $D$ up into two regions as follows:
$$I=\int_0^4 x\int_0^{\frac{3}{4}x}\,dy\,dx+\int_4^5 x\int_0^{\sqrt{25-x^2}}\,dy\,dx$$

$$I=\frac{3}{4}\int_0^4 x^2\,dx+\int_4^5 x\sqrt{25-x^2}\,dx$$

$$I=\frac{1}{4}\left[x^3\right]_0^4+\frac{1}{2}\int_0^9 u^{\frac{1}{2}}\,du$$

$$I=16+\frac{1}{3}9^{\frac{3}{2}}=16+9=25$$

Brilliantly done! Why two integrals for dydx and not dxdy?
 
Harpazo said:
Brilliantly done! Why two integrals for dydx and not dxdy?

Well, if you look at the region $D$, you can see that integrating in the original order (with horizontal segments), the left and right bounds of each segment are along the same curves all the way up from $y=0$ to $y=3$, but reversing the order and using vertical segments, we have the same bottom boundary from $x=0$ to $x=5$, but the upper boundary is one curve on $0\le x\le4$ and another curve from $4\le x\le 5$. And so to integrate in that order, we must use 2 integrals. :D
 
MarkFL said:
Well, if you look at the region $D$, you can see that integrating in the original order (with horizontal segments), the left and right bounds of each segment are along the same curves all the way up from $y=0$ to $y=3$, but reversing the order and using vertical segments, we have the same bottom boundary from $x=0$ to $x=5$, but the upper boundary is one curve on $0\le x\le4$ and another curve from $4\le x\le 5$. And so to integrate in that order, we must use 2 integrals. :D

I thank you for your great help with the set up. I will be posting more double integrals. I have trouble deciding if dxdy is better to use or dydx for certain questions.

In fact, I am having trouble finding the limits of integration for double integrals. I am also having trouble setting up the integrals. I will post 3 questions that involve setting up the double integrals. No need to integrate the double integral just the set up.
 
Harpazo said:
I thank you for your great help with the set up. I will be posting more double integrals. I have trouble deciding if dxdy is better to use or dydx for certain questions.

In fact, I am having trouble finding the limits of integration for double integrals. I am also having trouble setting up the integrals. I will post 3 questions that involve setting up the double integrals. No need to integrate the double integral just the set up.

To decide, do a sketch of your region, and draw in both horizontal and vertical strips. If one way can be done without the lower and upper boundaries changing, then choose that one.
 
Thank you everyone.
 

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